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;
char b[1000007];
vector<int> valid(4 * 1000007, 0), open(4 * 1000007, 0), closed(4 * 1000007, 0);
void preprocess(void) { return; }
void build(int ind, int l, int r) {
  if (r - l < 2) {
    if (b[l] == '(') {
      open[ind] = 1;
    } else {
      closed[ind] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  build(2 * ind, l, mid);
  build(2 * ind + 1, mid, r);
  int temp = min(open[2 * ind], closed[2 * ind + 1]);
  valid[ind] = valid[2 * ind] + valid[2 * ind + 1] + 2 * temp;
  open[ind] = open[2 * ind] + open[2 * ind + 1] - temp;
  closed[ind] = closed[ind * 2] + closed[ind * 2 + 1] - temp;
  return;
}
pair<int, pair<int, int> > solve(int x, int y, int ind, int l, int r) {
  if (l >= y || x >= r) {
    return pair<int, pair<int, int> >(0, pair<int, int>(0, 0));
  }
  if (x <= l && y >= r) {
    return pair<int, pair<int, int> >(valid[ind],
                                      pair<int, int>(open[ind], closed[ind]));
  }
  int mid = (l + r) / 2;
  pair<int, pair<int, int> > lc = solve(x, y, 2 * ind, l, mid);
  pair<int, pair<int, int> > rc = solve(x, y, 2 * ind + 1, mid, r);
  int temp = min(lc.second.first, rc.second.second);
  return pair<int, pair<int, int> >(
      lc.first + rc.first + 2 * temp,
      pair<int, int>(lc.second.first + rc.second.first - temp,
                     lc.second.second + rc.second.second - temp));
}
int main(void) {
  preprocess();
  int t;
  t = 1;
  for (int z = 1; z <= t; z++) {
    scanf("%s", b);
    int q;
    int n = strlen(b);
    scanf("%d", &q);
    build(1, 0, n);
    while (q--) {
      int l, r;
      scanf("%d", &l), scanf("%d", &r);
      l--, r;
      pair<int, pair<int, int> > ans = solve(l, r, 1, 0, n);
      printf("%d\n", ans.first);
    }
  }
  return 0;
}
