Sereja has a bracket sequence s1, s2, ..., sn, or, in other words, a string s of length n, consisting of characters "(" and ")".

Sereja needs to answer m queries, each of them is described by two integers li, ri (1 ≤ li ≤ ri ≤ n). The answer to the i-th query is the length of the maximum correct bracket subsequence of sequence sli, sli + 1, ..., sri. Help Sereja answer all queries.

You can find the definitions for a subsequence and a correct bracket sequence in the notes.

Input

The first line contains a sequence of characters s1, s2, ..., sn (1 ≤ n ≤ 106) without any spaces. Each character is either a "(" or a ")". The second line contains integer m (1 ≤ m ≤ 105) — the number of queries. Each of the next m lines contains a pair of integers. The i-th line contains integers li, ri (1 ≤ li ≤ ri ≤ n) — the description of the i-th query.

Output

Print the answer to each question on a single line. Print the answers in the order they go in the input.

Examples

Input

())(())(())(
7
1 1
2 3
1 2
1 12
8 12
5 11
2 10


Output

0
0
2
10
4
6
6

Note

A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is string x = sk1sk2... sk|x| (1 ≤ k1 < k2 < ... < k|x| ≤ |s|).

A correct bracket sequence is a bracket sequence that can be transformed into a correct aryphmetic expression by inserting characters "1" and "+" between the characters of the string. For example, bracket sequences "()()", "(())" are correct (the resulting expressions "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

For the third query required sequence will be «()».

For the fourth query required sequence will be «()(())(())».
#include <bits/stdc++.h>
int n, m;
char s[1000005];
struct stType {
  int sum[3];
};
stType st[1000005 << 2];
int min(int x, int y) { return x < y ? x : y; }
void build_st(int x, int l, int r) {
  if (l == r) {
    st[x].sum[s[l] == '('] = 1;
  } else {
    build_st((x << 1), l, ((l + r) >> 1));
    build_st((x << 1 | 1), ((l + r) >> 1) + 1, r);
    int tmp = min(st[(x << 1)].sum[1], st[(x << 1 | 1)].sum[0]);
    st[x].sum[2] = st[(x << 1)].sum[2] + st[(x << 1 | 1)].sum[2] + tmp * 2;
    st[x].sum[0] = st[(x << 1)].sum[0] + st[(x << 1 | 1)].sum[0] - tmp;
    st[x].sum[1] = st[(x << 1)].sum[1] + st[(x << 1 | 1)].sum[1] - tmp;
  }
}
void standing_by() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  build_st(1, 1, n);
}
stType query_st(int x, int l, int r, int L, int R) {
  stType tmp;
  if (r < L || l > R) {
    tmp.sum[0] = tmp.sum[1] = tmp.sum[2] = 0;
    return tmp;
  } else if (l >= L && r <= R)
    return st[x];
  else {
    stType tmpL, tmpR;
    int tmpM;
    tmpL = query_st((x << 1), l, ((l + r) >> 1), L, R);
    tmpR = query_st((x << 1 | 1), ((l + r) >> 1) + 1, r, L, R);
    tmpM = min(tmpL.sum[1], tmpR.sum[0]);
    tmp.sum[2] = tmpL.sum[2] + tmpR.sum[2] + tmpM * 2;
    tmp.sum[0] = tmpL.sum[0] + tmpR.sum[0] - tmpM;
    tmp.sum[1] = tmpL.sum[1] + tmpR.sum[1] - tmpM;
    return tmp;
  }
}
void complete() {
  int i, l, r;
  stType ans;
  scanf("%d", &m);
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &l, &r);
    ans = query_st(1, 1, n, l, r);
    printf("%d\n", ans.sum[2]);
  }
}
int main() {
  standing_by();
  complete();
  return 0;
}
