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;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(long long N, int pos) { return (bool)(N & (1LL << pos)); }
string arr, s;
int m, tot[4 * 1000001], op[4 * 1000001], cl[4 * 1000001];
struct node {
  int a, b, c;
};
void build(int id, int l, int r) {
  if (l == r) {
    if (arr[l] == '(')
      op[id] = 1;
    else
      cl[id] = 1;
    return;
  }
  int mid = (l + r) / 2;
  build(id * 2, l, mid);
  build(id * 2 + 1, mid + 1, r);
  int mn = min(op[id * 2], cl[id * 2 + 1]);
  op[id] = op[id * 2] + op[id * 2 + 1] - mn;
  cl[id] = cl[id * 2] + cl[id * 2 + 1] - mn;
  tot[id] = tot[id * 2] + tot[id * 2 + 1] + mn;
}
node query(int id, int l, int r, int x, int y) {
  if (r < x || l > y) {
    node ret;
    ret.a = ret.b = ret.c = 0;
    return ret;
  }
  if (l >= x && r <= y) {
    node ret;
    ret.a = op[id];
    ret.b = cl[id];
    ret.c = tot[id];
    return ret;
  }
  int mid = (l + r) / 2;
  node ret1 = query(id * 2, l, mid, x, y);
  node ret2 = query(id * 2 + 1, mid + 1, r, x, y);
  node ret;
  int mn = min(ret1.a, ret2.b);
  ret.a = ret1.a + ret2.a - mn;
  ret.b = ret1.b + ret2.b - mn;
  ret.c = ret1.c + ret2.c + mn;
  return ret;
}
int main() {
  cin >> s;
  arr = " ";
  arr += s;
  build(1, 1, s.size());
  scanf("%d", &m);
  int x, y;
  while (m--) {
    scanf("%d %d", &x, &y);
    printf("%d\n", query(1, 1, s.size(), x, y).c * 2);
  }
  return 0;
}
