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;
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 5;
struct SegTree {
  int lf, rt;
  int a, b, c;
} st[maxn << 2];
char s[maxn];
void PushUp(int x) {
  SegTree &now = st[x];
  SegTree &lc = st[x << 1];
  SegTree &rc = st[x << 1 | 1];
  int ok = min(lc.a, rc.b);
  now.a = lc.a + rc.a - ok;
  now.b = lc.b + rc.b - ok;
  now.c = lc.c + rc.c + ok;
}
void Build(int x, int lf, int rt) {
  SegTree &now = st[x];
  now.lf = lf;
  now.rt = rt;
  now.a = now.b = now.c = 0;
  if (lf == rt) {
    if (s[lf] == '(')
      ++now.a;
    else
      ++now.b;
    return;
  }
  int mid = (lf + rt) >> 1;
  Build(x << 1, lf, mid);
  Build(x << 1 | 1, mid + 1, rt);
  PushUp(x);
}
int qa, qb, qc;
void Query(int x, int lf, int rt) {
  SegTree &now = st[x];
  if (now.lf >= lf && now.rt <= rt) {
    int ok = min(qa, now.b);
    qa = qa + now.a - ok;
    qb = qb + now.b - ok;
    qc = qc + now.c + ok;
    return;
  }
  int mid = (now.lf + now.rt) >> 1;
  if (lf <= mid) Query(x << 1, lf, rt);
  if (rt > mid) Query(x << 1 | 1, lf, rt);
}
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  Build(1, 1, n);
  int m;
  scanf("%d", &m);
  for (int im = 1; im <= m; ++im) {
    int l, r;
    scanf("%d%d", &l, &r);
    qa = qb = qc = 0;
    Query(1, l, r);
    printf("%d\n", qc * 2);
  }
  return 0;
}
