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 c[1000017], ans[100017];
char s[1000017];
int m, len;
stack<int> st;
struct node {
  int l, r, id;
} q[100017];
int lowbit(int x) { return x & (-x); }
void update(int x, int k) {
  while (x <= len) {
    c[x] += k;
    x += lowbit(x);
  }
}
int sum(int x) {
  int ans = 0;
  while (x) {
    ans += c[x];
    x -= lowbit(x);
  }
  return ans;
}
inline bool cmp(node x, node y) { return x.r < y.r; }
int main() {
  while (~scanf("%s", s)) {
    int pos = 1;
    memset(c, 0, sizeof(c));
    while (!st.empty()) st.pop();
    len = strlen(s);
    scanf("%d", &m);
    for (int i = 0; i < m; ++i) {
      scanf("%d%d", &q[i].l, &q[i].r);
      q[i].id = i;
    }
    sort(q, q + m, cmp);
    for (int i = 0; i < m; ++i) {
      for (int j = pos; j <= q[i].r; ++j) {
        if (s[j - 1] == '(')
          st.push(j);
        else if (s[j - 1] == ')') {
          if (!st.empty()) {
            int tt = st.top();
            update(tt, 2);
            st.pop();
          }
        }
      }
      pos = q[i].r + 1;
      ans[q[i].id] = sum(q[i].r) - sum(q[i].l - 1);
    }
    for (int i = 0; i < m; ++i) printf("%d\n", ans[i]);
  }
  return 0;
}
