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 N = int(1e6) + 10;
int n, m;
char second[N];
int l, r;
int a[N], b[N];
struct Seg {
  int a, b, c;
  Seg() {
    a = 0;
    b = 0;
    c = 0;
  }
  Seg(int x, int y, int z) {
    a = x;
    b = y;
    c = z;
  }
  Seg operator+(Seg x) {
    int t = min(b, x.c);
    return Seg(x.a + a + t, b + x.b - t, c + x.c - t);
  }
} t[4 * N];
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = Seg(0, a[tl], b[tl]);
    return;
  }
  int tm = (tl + tr) / 2;
  build(v * 2, tl, tm);
  build(v * 2 + 1, tm + 1, tr);
  t[v] = t[2 * v] + t[2 * v + 1];
}
Seg get(int v, int tl, int tr, int l, int r) {
  if (l > tr || tl > r) return Seg(0, 0, 0);
  if (l <= tl && tr <= r) return t[v];
  int tm = (tl + tr) / 2;
  return get(v * 2, tl, tm, l, r) + get(v * 2 + 1, tm + 1, tr, l, r);
}
int main() {
  scanf("%s\n", second + 1);
  n = strlen(second + 1);
  for (int i = 1; i <= n; i++) {
    if (second[i] == '(')
      a[i] = 1;
    else
      b[i] = 1;
  }
  build(1, 1, n);
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d", &l, &r);
    printf("%d\n", 2 * get(1, 1, n, l, r).a);
  }
  return 0;
}
