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 maxn = -1e18;
long long minn = 1e18;
long long mod = 1e9 + 7;
const int maxx = 1e6 + 5;
struct viet {
  long long o, c, a;
};
viet st[maxx * 4];
string s;
void upd(long long k, long long l, long long r, long long x) {
  if (x < l || x > r) return;
  if (l == r) {
    if (s[x] == '(') {
      st[k].o = 1;
      st[k].c = 0;
      st[k].a = 0;
    } else {
      st[k].o = 0;
      st[k].c = 1;
      st[k].a = 0;
    }
    return;
  }
  long long mid = (l + r) / 2;
  upd(k * 2, l, mid, x);
  upd(k * 2 + 1, mid + 1, r, x);
  long long t = min(st[k * 2].o, st[k * 2 + 1].c);
  st[k].a = st[k * 2].a + st[k * 2 + 1].a + t * 2;
  st[k].o = st[k * 2].o + st[k * 2 + 1].o - t;
  st[k].c = st[k * 2].c + st[k * 2 + 1].c - t;
}
viet get(long long k, long long l, long long r, long long x, long long y) {
  if (l > y || r < x) return {0, 0, 0};
  viet ret;
  if (l >= x && y >= r) {
    ret.a = st[k].a;
    ret.o = st[k].o;
    ret.c = st[k].c;
    return ret;
  }
  long long mid = (l + r) / 2;
  viet left = get(k * 2, l, mid, x, y);
  viet right = get(k * 2 + 1, mid + 1, r, x, y);
  long long t = min(left.o, right.c);
  ret.a = left.a + right.a + t * 2;
  ret.o = left.o + right.o - t;
  ret.c = left.c + right.c - t;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  long long n = s.length();
  s = " " + s;
  for (int i = 1; i <= n; i++) {
    upd(1, 1, n, i);
  }
  long long m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    long long l, r;
    cin >> l >> r;
    cout << get(1, 1, n, l, r).a << '\n';
  }
}
