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 State {
  int pair, open, close;
};
vector<State> precalc;
const int BLOCK = 1000;
State compute(const string &s) {
  State result = {0, 0, 0};
  for (int i = 0; i < (int)(s.length()); i++) {
    if (s[i] == '(') {
      result.open++;
    } else {
      if (result.open > 0) {
        result.open--;
        result.pair++;
      } else {
        result.close++;
      }
    }
  }
  return result;
}
void doPrecalc(const string &s) {
  precalc.clear();
  for (int i = 0; i < s.length(); i += BLOCK) {
    string t = s.substr(i, BLOCK);
    precalc.push_back(compute(t));
  }
}
int merge(const vector<State> &s) {
  State result = {0, 0, 0};
  for (int i = 0; i < (int)(s.size()); i++) {
    result.pair += s[i].pair;
    int p = min(result.open, s[i].close);
    result.pair += p;
    result.open -= p;
    result.open += s[i].open;
  }
  return result.pair * 2;
}
int main() {
  static char buf[1 << 20];
  while (scanf("%s", buf) != -1) {
    string s(buf);
    doPrecalc(s);
    int n;
    scanf("%d", &n);
    while (n--) {
      int src, dst;
      scanf("%d%d", &src, &dst);
      src--;
      vector<State> v;
      while (src < dst) {
        int srcBlockBegin = src / BLOCK * BLOCK;
        int srcBlockEnd = srcBlockBegin + BLOCK;
        int block = src / BLOCK;
        if (src == srcBlockBegin && srcBlockEnd <= dst) {
          v.push_back(precalc[block]);
          src = srcBlockEnd;
        } else {
          int end = min(srcBlockEnd, dst);
          v.push_back(compute(s.substr(src, end - src)));
          src = end;
        }
      }
      printf("%d\n", merge(v));
    }
  }
}
