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;
struct Node {
  int number;
  int open;
  int close;
};
Node construct(char s[], Node *st, int si, int se, int node) {
  if (si == se) {
    st[node].number = 0;
    if (s[si] == '(') {
      st[node].open = 1;
      st[node].close = 0;
    } else {
      st[node].open = 0;
      st[node].close = 1;
    }
    return st[node];
  }
  int mid = si + (se - si) / 2;
  Node left = construct(s, st, si, mid, 2 * node + 1);
  Node right = construct(s, st, mid + 1, se, 2 * node + 2);
  int add = (left.open) < (right.close) ? left.open : right.close;
  st[node].number = left.number + right.number + add;
  st[node].open = left.open - add + right.open;
  st[node].close = left.close + right.close - add;
  return st[node];
}
Node *segTree(char s[], int n) {
  int x = ceil(log2(n));
  int maxSize = pow(2, x + 1) - 1;
  Node *st = new Node[maxSize];
  construct(s, st, 0, n - 1, 0);
  return st;
}
Node *ansFind(Node *st, int si, int se, int qi, int qe, int node) {
  if (qi <= si && qe >= se) return &st[node];
  Node *ans = new Node;
  if (si > qe || se < qi) {
    ans->number = 0;
    ans->open = 0;
    ans->close = 0;
    return ans;
  }
  int mid = si + (se - si) / 2;
  Node *left = ansFind(st, si, mid, qi, qe, 2 * node + 1);
  Node *right = ansFind(st, mid + 1, se, qi, qe, 2 * node + 2);
  int add = (left->open) < (right->close) ? left->open : right->close;
  ans->number = left->number + right->number + add;
  ans->open = left->open - add + right->open;
  ans->close = left->close + right->close - add;
  return ans;
}
int ans(Node *st, int n, int qi, int qe) {
  Node *nod = ansFind(st, 0, n - 1, qi, qe, 0);
  return nod->number;
}
int main() {
  ios_base::sync_with_stdio(0);
  char s[1000001];
  cin >> s;
  int n = strlen(s);
  Node *st = segTree(s, n);
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int qi, qe;
    cin >> qi >> qe;
    cout << ans(st, n, qi - 1, qe - 1) * 2 << "\n";
  }
  return 0;
}
