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 = 1000006;
struct node {
  int open, close, maxtrue;
};
int l, r;
int n, m;
node rmq[maxn * 4];
char a[maxn];
void build(int v, int l, int r) {
  if (l + 1 == r)
    if (a[l] == ')')
      rmq[v].close = 1;
    else
      rmq[v].open = 1;
  else {
    int m = (l + r) / 2;
    build(v * 2 + 1, l, m);
    build(v * 2 + 2, m, r);
    int t = min(rmq[v * 2 + 1].open, rmq[v * 2 + 2].close);
    rmq[v].maxtrue = rmq[v * 2 + 1].maxtrue + rmq[v * 2 + 2].maxtrue + t;
    rmq[v].open = rmq[v * 2 + 1].open + rmq[v * 2 + 2].open - t;
    rmq[v].close = rmq[v * 2 + 1].close + rmq[v * 2 + 2].close - t;
  }
}
node get(int v, int tl, int tr, int l, int r) {
  if ((tl >= r) || (tr <= l)) return {0, 0, 0};
  if ((tl >= l) && (tr <= r)) return rmq[v];
  int m = (tr + tl) / 2;
  node q;
  node q1 = get(v * 2 + 1, tl, m, l, r);
  node q2 = get(v * 2 + 2, m, tr, l, r);
  int t = min(q1.open, q2.close);
  q.maxtrue = q1.maxtrue + q2.maxtrue + t;
  q.open = q1.open + q2.open - t;
  q.close = q1.close + q2.close - t;
  return q;
}
int main() {
  scanf(" %s", a);
  n = strlen(a);
  cin >> m;
  build(0, 0, n);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &l, &r);
    l--;
    printf("%d\n", 2 * get(0, 0, n, l, r).maxtrue);
  }
  return 0;
}
