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;
long long t, m, k, ans, x, y, cnt;
long long inf = 1e18;
long long mn = 1e9, mx = -inf;
string s;
const int N = 1000100;
int n;
char a[N];
struct node {
  int op, cl, crt;
};
node seg[N << 2];
node emp;
node mrg(node a, node b) {
  node rt;
  rt.crt = 2 * min(a.op, b.cl) + a.crt + b.crt;
  rt.cl = a.cl + (b.cl - min(a.op, b.cl));
  rt.op = b.op + (a.op - min(a.op, b.cl));
  return rt;
}
void build(int nd, int L, int R) {
  if (L == R) {
    seg[nd].crt = 0;
    if (a[L] == ')') seg[nd].cl = 1;
    if (a[L] == '(') seg[nd].op = 1;
    return;
  }
  int mid = (L + R) >> 1;
  build(nd << 1, L, mid);
  build(nd << 1 | 1, mid + 1, R);
  seg[nd] = mrg(seg[nd << 1], seg[nd << 1 | 1]);
}
node query(int nd, int L, int R, int from, int to) {
  if (from <= L && R <= to) return seg[nd];
  if (from > R || to < L) return emp;
  int mid = (L + R) >> 1;
  node leftAns = query(nd << 1, L, mid, from, to);
  node rightAns = query(nd << 1 | 1, mid + 1, R, from, to);
  return mrg(leftAns, rightAns);
}
int main() {
  cin >> s >> m;
  n = s.length();
  for (int i = 1; i <= n; i++) a[i] = s[i - 1];
  build(1, 1, n);
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    node dd = query(1, 1, n, x, y);
    cout << dd.crt << endl;
  }
}
