You've got an array *a*, consisting of *n* integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>. You are allowed to perform two operations on this array:
 1.  Calculate the sum of current array elements on the segment [*l*, *r*], that is, count value *a*<sub class="lower-index">*l*</sub> + *a*<sub class="lower-index">*l* + 1</sub> + ... + *a*<sub class="lower-index">*r*</sub>. 1.  Apply the xor operation with a given number *x* to each array element on the segment [*l*, *r*], that is, execute <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/2ef04d094daf2dcff9a998a558bd9cab93bbd012.png" style="max-width: 100.0%;max-height: 100.0%;"/>. This operation changes exactly *r* - *l* + 1 array elements. 
Expression <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b364f2e04c665b78b924ec10666327a4ef4635bc.png" style="max-width: 100.0%;max-height: 100.0%;"/> means applying bitwise xor operation to numbers *x* and *y*. The given operation exists in all modern programming languages, for example in language C++ and Java it is marked as "^", in Pascal — as "xor".

You've got a list of *m* operations of the indicated type. Your task is to perform all given operations, for each sum query you should print the result you get.
The first line contains integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>) — the size of the array. The second line contains space-separated integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (0 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">6</sup>) — the original array.

The third line contains integer *m* (1 ≤ *m* ≤ 5·10<sup class="upper-index">4</sup>) — the number of operations with the array. The *i*-th of the following *m* lines first contains an integer *t*<sub class="lower-index">*i*</sub> (1 ≤ *t*<sub class="lower-index">*i*</sub> ≤ 2) — the type of the *i*-th query. If *t*<sub class="lower-index">*i*</sub> = 1, then this is the query of the sum, if *t*<sub class="lower-index">*i*</sub> = 2, then this is the query to change array elements. If the *i*-th operation is of type 1, then next follow two integers *l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub> (1 ≤ *l*<sub class="lower-index">*i*</sub> ≤ *r*<sub class="lower-index">*i*</sub> ≤ *n*). If the *i*-th operation is of type 2, then next follow three integers *l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub>, *x*<sub class="lower-index">*i*</sub> (1 ≤ *l*<sub class="lower-index">*i*</sub> ≤ *r*<sub class="lower-index">*i*</sub> ≤ *n*, 1 ≤ *x*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">6</sup>). The numbers on the lines are separated by single spaces.
import sys
n = int(input())
a = list(map(int,sys.stdin.readline().strip().split()))
m = int(input())
for _ in range(m):
    q = list(map(int,sys.stdin.readline().strip().split()))
    if(q[0]==1):
        ans = 0
        for i in range(q[1]-1,q[2]):
            ans+=a[i]
        print(ans)
    else:
        for i in range(q[1]-1,q[2]):
            a[i]^=q[3]