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;
inline long long rd() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
const int N = 1e6 + 7;
char c[N];
int sum[N << 2], ch[N << 2][2], n;
void update(int p) {
  int ls = p << 1, rs = p << 1 | 1;
  sum[p] = min(ch[ls][0], ch[rs][1]);
  ch[p][0] = ch[ls][0] + ch[rs][0] - sum[p];
  ch[p][1] = ch[ls][1] + ch[rs][1] - sum[p];
  sum[p] += sum[ls] + sum[rs];
}
void biu(int p, int l, int r) {
  if (l == r) {
    if (c[l] == '(')
      ch[p][0] = 1;
    else
      ch[p][1] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  biu(p << 1, l, mid);
  biu(p << 1 | 1, mid + 1, r);
  update(p);
}
int query(int p, int l, int r, int x, int y, int &L, int &R) {
  if (x <= l && y >= r) {
    L += ch[p][0], R += ch[p][1];
    return sum[p];
  }
  int mid = (l + r) >> 1, ls = 0, rs = 0, a, b, ans;
  if (y <= mid)
    return query(p << 1, l, mid, x, y, L, R);
  else if (x > mid)
    return query(p << 1 | 1, mid + 1, r, x, y, L, R);
  else {
    a = query(p << 1, l, mid, x, y, ls, R);
    b = query(p << 1 | 1, mid + 1, r, x, y, L, rs);
    ans = min(ls, rs);
    L = L + ls - ans;
    R = R + rs - ans;
    return ans + a + b;
  }
}
int main() {
  scanf("%s", c + 1);
  n = strlen(c + 1);
  biu(1, 1, n);
  int m = rd(), a, b, x, y;
  while (m--) {
    a = rd(), b = rd(), printf("%d\n", query(1, 1, n, a, b, x, y) * 2);
  }
}
