Once Petya read a problem about a bracket sequence. He gave it much thought but didn't find a solution. Today you will face it.

You are given string *s*. It represents a correct bracket sequence. A correct bracket sequence is the sequence of opening ("(") and closing (")") brackets, such that it is possible to obtain a correct mathematical expression from it, inserting numbers and operators between the brackets. For example, such sequences as "(())()" and "()" are correct bracket sequences and such sequences as ")()" and "(()" are not.

In a correct bracket sequence each bracket corresponds to the matching bracket (an opening bracket corresponds to the matching closing bracket and vice versa). For example, in a bracket sequence shown of the figure below, the third bracket corresponds to the matching sixth one and the fifth bracket corresponds to the fourth one.

You are allowed to color some brackets in the bracket sequence so as all three conditions are fulfilled: 
 -  Each bracket is either not colored any color, or is colored red, or is colored blue. -  For any pair of matching brackets exactly one of them is colored. In other words, for any bracket the following is true: either it or the matching bracket that corresponds to it is colored. -  No two neighboring colored brackets have the same color. 
Find the number of different ways to color the bracket sequence. The ways should meet the above-given conditions. Two ways of coloring are considered different if they differ in the color of at least one bracket. As the result can be quite large, print it modulo 1000000007 (10<sup class="upper-index">9</sup> + 7).
The first line contains the single string *s* (2 ≤ |*s*| ≤ 700) which represents a correct bracket sequence.
(())
12
(()())
40
()
4
((()))
36
()(())
42
()()()
48
(())(())
126
()()()()()()()()()()()(())
9085632
()(())()((()))
4428
()()(())()(())
5040
()()()()()()()()()()()()()()()()
411525376
(()()())
136
()(()())()
480
(())()(())()
1476
()()(()())(())()()()
195840
()()()((((())))())()()()()()((()))()()(())()(((())))()(()())((())())((()())(((((()()()())()()())))))
932124942
((()(((((()(()(())))()((((((((())))()(((((())()((((())())(()(()(())())((()))()((()))))))))))))))))))))
90824888
((()))((())())((()()))()(())(()())(())()()()((()(((()())))()())()((((()((()((())))(())(()(())())))((()())()()()((())()))()(())(())))()(((((()())))))))
100627207
()(((()((((()())))())(())(((((()(()()))))()()))((())))()())((())))(())()((()())())()(()(()())())(()())()(()(((((()))()((()()(())()(())(()((()((()))))()(())()()(()()()((((()())()))))()(((()(((((()()((((())(())))()())(()))(((())((()())(()))())(((()()()(()(())())())(()()()))))())))()((()(()()(()))())((())(()()()(())()))()()(((())))((()))(()((()(((()))((((()())))())(((())()(()((())))))))))))))))))))))
306199947
(())(((((()()()()())(())))(()()((()(()(((((())(()())))())(()()(()((())()(()()))))))(())()())))()((()()())))()()(()(())())()())()(())(((((()(()()(((()())()))((())((((()()()))())(((())(((())))))))))))))
270087235
()()()((()))(())(((())()(())(())))()()(((()((()()()))(()()(())(())))(()()((()((())(()()(()(())))))))(((())()((((()())))()(((()()())))()))()())))()(()(()())((()((()))))())(((((()())()((((()))(((((()())()))(((()()()((((((()()(())(()))((()(()(()((()((((()(((()(()()(()()((((()))()()()(()((((()(((())(((()()()(())()))((()()()(()))))())()))))(((((((()))())))(((()(()())(())))())))((((())(())())(((()()()))((()()))())(()))(())((()(()))(()()((()(()((()(())(()))()()))))))))))))))))))))))))))))))))))))))))))
461776571
()()(((((()((()(())()(()))(()(()(()(()(())(())(())(()(()((())))()))())((()((()(()(((()(()))()(()())(()()()()(((((()(((()))((((())())(((()((((()((((((())())))()))))))))(())())))(((()((()))))((())(()()))()(()(()((()())())()))))((()))))()((())())(()())()())))())())())())()((()((())((()()())()())())()(())()))(()(()))())))(()()()())()())))))))((((()())))((((()()()))())((()(())))))()((()(((())()()()(()()()()()))))(((()())()))()()(((())(()())(()()))))))
66338682
(()())()()()((((()(()()(())()((())(((()((()()(()))()))()()))))()(()(())(()))))))
639345575
()((()))((((()((())((()()((((()))()()((())((()(((((()(()))((())()))((((())()(()(()))()))))))))))))))))))
391997323
(((((()())))))()()()()()(())()()()((()()))()()()()()(((()(())))())(((()())))
422789312
((()((()))()((()(()))())))()((()())())()(()())((()))(()())(())()(())()(())(())((()()))((()))()()()()(())()
140121189
()()
14
from functools import lru_cache, reduce  # cache,
s = input()

MOD = 1000000007

mat = [0] * len(s)  # 存储了对应右括号的下标
stack = []
for i in range(len(s)):
    if s[i] == '(':
        stack.append(i)
    else:
        mat[stack.pop()] = i
# print(mat)


@lru_cache(None)
def dp(l, r, cl, cr):
    """左端点，右端点，左端点的颜色，右端点的颜色"""
    if mat[l] == r:  # 左右端点匹配，说明是一个区间
        if (cl > 0 and cr > 0) or cl + cr == 0:  # 不合法的情况，只且只有一边有颜色
            return 0
        if l == r - 1:  # 特判 () 这种情况，在固定cl,cr的情况下，是一种情况
            return 1
        ans = 0
        for i in range(3):
            for j in range(3):
                if (0 < cl == i) or (0 < cr == j):  # i,j作为新的cl，cr，不能和之前的颜色相同（若之前是0未涂色，则可以相同）
                    continue
                ans += dp(l + 1, r - 1, i, j)  # 开始递归，记忆化搜索。。
        ans %= MOD
        return ans
    else:  # 当前区间为多个有效区间的组合
        rr = mat[l]  # 当前左端点l实际匹配的右端点
        ans = 0
        for i in range(3):
            for j in range(3):
                if i != j or (i + j == 0):  # ij相邻，颜色不能相同，或者都是未涂色
                    ans += dp(l, rr, cl, i) * dp(rr + 1, r, j, cr)
        ans %= MOD
        return ans


ans = 0
for i in range(3):
    for j in range(3):
        ans += dp(0, len(s)-1, i, j)

print(ans % MOD)