The bear has a string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (record |*s*| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices *i*, *j* (1 ≤ *i* ≤ *j* ≤ |*s*|), that string *x*(*i*, *j*) = *s*<sub class="lower-index">*i*</sub>*s*<sub class="lower-index">*i* + 1</sub>... *s*<sub class="lower-index">*j*</sub> contains at least one string "bear" as a substring.

String *x*(*i*, *j*) contains string "bear", if there is such index *k* (*i* ≤ *k* ≤ *j* - 3), that *s*<sub class="lower-index">*k*</sub> = *b*, *s*<sub class="lower-index">*k* + 1</sub> = *e*, *s*<sub class="lower-index">*k* + 2</sub> = *a*, *s*<sub class="lower-index">*k* + 3</sub> = *r*.

Help the bear cope with the given problem.
The first line contains a non-empty string *s* (1 ≤ |*s*| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
def count_bear_substrings(s):
    n = len(s)
    dp = [[0] * n for _ in range(n)]

    for i in range(n-3, -1, -1):
        for j in range(i+3, n):
            if s[i] == 'b' and s[i+1] == 'e' and s[i+2] == 'a' and s[i+3] == 'r':
                dp[i][j] = dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1] + 1
            elif s[i] == 'b':
                dp[i][j] = dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1]
            else:
                dp[i][j] = dp[i+1][j] + dp[i][j-1]

    return dp[0][n-1]
s = input().strip()
print(count_bear_substrings(s))
