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 dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, -1, 1};
struct node {
  int a, f, q;
  node(){};
  node(int aa, int bb, int cc) {
    a = aa;
    f = bb;
    q = cc;
  }
  node operator+(node x) const {
    node foo;
    int tot = min(a, x.f);
    foo.q = q + x.q + 2 * tot;
    foo.a = a + x.a - tot;
    foo.f = f + x.f - tot;
    return foo;
  }
};
char s[1000100];
int m, n;
node tree[4000100];
void build(int no, int l, int r) {
  if (l == r) {
    if (s[l] == '(')
      tree[no] = node(1, 0, 0);
    else
      tree[no] = node(0, 1, 0);
    return;
  }
  int nxt = no * 2;
  int mid = (l + r) / 2;
  build(nxt, l, mid);
  build(nxt + 1, mid + 1, r);
  tree[no] = tree[nxt] + tree[nxt + 1];
}
node query(int no, int l, int r, int i, int j) {
  if (r < i || l > j) return node(0, 0, 0);
  if (l >= i && r <= j) return tree[no];
  int nxt = (no << 1);
  int mid = (l + r) >> 1;
  return query(nxt, l, mid, i, j) + query(nxt + 1, mid + 1, r, i, j);
}
int main() {
  scanf(" %s %d", s, &m);
  n = strlen(s);
  build(1, 0, n - 1);
  while (m--) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    printf("%d\n", query(1, 0, n - 1, a, b).q);
  }
  return 0;
}
