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 n, m, tt, cnt, k, x, y, A[1000025], start, finish;
string cad;
struct node {
  int open, close, len;
  node(){};
  node(int o, int c, int l) {
    open = o;
    close = c;
    len = l;
  }
} T[4 * 1000025];
void create(int nodo, int ini, int fin) {
  if (ini == fin) {
    T[nodo].open = 0;
    T[nodo].close = 0;
    T[nodo].len = 0;
    if (cad[ini - 1] == '(')
      T[nodo].open = 1;
    else
      T[nodo].close = 1;
    return;
  }
  int med = (ini + fin) / 2;
  create(2 * nodo, ini, med);
  create(2 * nodo + 1, med + 1, fin);
  int adds = min(T[2 * nodo].open, T[2 * nodo + 1].close);
  T[nodo].len = T[2 * nodo].len + T[2 * nodo + 1].len + 2 * adds;
  T[nodo].open = T[2 * nodo].open + T[2 * nodo + 1].open - adds;
  T[nodo].close = T[2 * nodo].close + T[2 * nodo + 1].close - adds;
}
node query(int nodo, int ini, int fin) {
  if (fin < start || ini > finish) return node(0, 0, 0);
  if (ini >= start && fin <= finish) return T[nodo];
  int med = (ini + fin) / 2;
  node aa = query(2 * nodo, ini, med);
  node bb = query(2 * nodo + 1, med + 1, fin);
  node cc = node(0, 0, 0);
  int adds = min(aa.open, bb.close);
  cc.len = aa.len + bb.len + 2 * adds;
  cc.open = aa.open + bb.open - adds;
  cc.close = aa.close + bb.close - adds;
  return cc;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> cad;
  n = cad.length();
  create(1, 1, n);
  cin >> m;
  while (m--) {
    cin >> start >> finish;
    node r = query(1, 1, n);
    cout << r.len << endl;
  }
}
