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 const N = 1000000;
struct Q {
  int a, b, c;
  Q(int x = 0, int y = 0, int z = 0) {
    a = x;
    b = y;
    c = z;
  }
} seg[4 * N + 1];
char g[N + 1];
void ad(int u, int s, int e) {
  if (s == e) {
    seg[u].c = (g[s] == '(');
    seg[u].b += (g[s] == ')');
    return;
  }
  ad(u * 2, s, (s + e) / 2);
  ad(u * 2 + 1, (s + e) / 2 + 1, e);
  Q x = seg[u * 2], y = seg[u * 2 + 1];
  Q &cur = seg[u];
  int t = min(x.c, y.b);
  cur = x.a + y.a + t;
  x.c -= t, y.b -= t;
  cur.b = x.b + y.b;
  cur.c = x.c + y.c;
}
int l, r;
Q get(int u, int s, int e) {
  if (s > r || e < l) return Q(0, 0, 0);
  if (s >= l && e <= r) {
    return seg[u];
  }
  Q x = get(u * 2, s, (s + e) / 2), y = get(u * 2 + 1, (s + e) / 2 + 1, e);
  Q cur = seg[u];
  int t = min(x.c, y.b);
  cur = x.a + y.a + t;
  x.c -= t, y.b -= t;
  cur.b = x.b + y.b;
  cur.c = x.c + y.c;
  return cur;
}
int main() {
  int m;
  scanf("%s", g);
  int n = strlen(g);
  ad(1, 0, n - 1);
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &l, &r);
    --l, --r;
    Q an = get(1, 0, n - 1);
    printf("%d\n", an.a * 2);
  }
}
