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>
using namespace std;
const int maxn = 1000000 + 10;
int sum[maxn << 2], ll[maxn << 2], rr[maxn << 2];
char s[maxn];
void PushUp(int rt) {
  int tmp = min(ll[rt << 1], rr[rt << 1 | 1]);
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1] + tmp * 2;
  ll[rt] = ll[rt << 1] + ll[rt << 1 | 1] - tmp;
  rr[rt] = rr[rt << 1] + rr[rt << 1 | 1] - tmp;
}
void build(int l, int r, int rt) {
  if (l == r) {
    ll[rt] = rr[rt] = sum[rt] = 0;
    if (s[l - 1] == '(')
      ll[rt] = 1;
    else
      rr[rt] = 1;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  PushUp(rt);
}
int Query(int L, int R, int l, int r, int rt, int &lls, int &rrs) {
  if (l >= L && r <= R) {
    lls = ll[rt];
    rrs = rr[rt];
    return sum[rt];
  }
  int m = (l + r) >> 1;
  if (m >= R)
    return Query(L, R, l, m, rt << 1, lls, rrs);
  else if (m < L)
    return Query(L, R, m + 1, r, rt << 1 | 1, lls, rrs);
  else {
    int sum1, sum2, ll2, rr2, tmp;
    sum1 = Query(L, R, l, m, rt << 1, lls, rrs);
    sum2 = Query(L, R, m + 1, r, rt << 1 | 1, ll2, rr2);
    tmp = min(lls, rr2);
    lls = lls + ll2 - tmp;
    rrs = rrs + rr2 - tmp;
    return sum1 + sum2 + tmp * 2;
  }
}
int main() {
  int n;
  scanf("%s", s);
  n = strlen(s);
  build(1, n, 1);
  int m, l, r, ll, rr;
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d", &l, &r);
    int ans = Query(l, r, 1, n, 1, ll, rr);
    printf("%d\n", ans);
  }
  return 0;
}
