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 = 1000010;
int lsum[maxn << 2], rsum[maxn << 2], msum[maxn << 2];
char str[maxn];
struct point {
  int l, r, m;
  point(int l = 0, int r = 0, int m = 0) : l(l), r(r), m(m) {}
};
void pushup(int o) {
  lsum[o] = lsum[2 * o] + lsum[2 * o + 1];
  rsum[o] = rsum[2 * o] + rsum[2 * o + 1];
  msum[o] = msum[2 * o] + msum[2 * o + 1];
  msum[o] += min(lsum[2 * o] - msum[2 * o], rsum[2 * o + 1] - msum[2 * o + 1]);
}
void build(int l, int r, int o) {
  if (l == r) {
    lsum[o] = rsum[o] = msum[o] = 0;
    if (str[l] == '(') lsum[o] = 1;
    if (str[l] == ')') rsum[o] = 1;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, 2 * o);
  build(m + 1, r, 2 * o + 1);
  pushup(o);
}
point query(int L, int R, int l, int r, int o) {
  if (L <= l && R >= r) {
    return point(lsum[o], rsum[o], msum[o]);
  }
  int m = (l + r) >> 1;
  if (L > m) return query(L, R, m + 1, r, 2 * o + 1);
  if (R <= m) return query(L, R, l, m, 2 * o);
  point p1 = query(L, R, l, m, 2 * o), p2 = query(L, R, m + 1, r, 2 * o + 1);
  point p;
  p.l = p1.l + p2.l;
  p.r = p1.r + p2.r;
  p.m = p1.m + p2.m + min(p1.l - p1.m, p2.r - p2.m);
  return p;
}
int main() {
  scanf("%s", str);
  int n = strlen(str);
  build(0, n - 1, 1);
  int m;
  scanf("%d", &m);
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    point p = query(l, r, 0, n - 1, 1);
    printf("%d\n", 2 * p.m);
  }
  return 0;
}
