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 maxn = 1e6 + 10;
char s[maxn];
struct node {
  int l, r;
  int a, b, ans;
} Tree[maxn << 2];
void pushup(int rt) {
  int tt = min(Tree[rt * 2].a, Tree[rt * 2 + 1].b);
  Tree[rt].a = Tree[rt * 2].a + Tree[rt * 2 + 1].a - tt;
  Tree[rt].b = Tree[rt * 2].b + Tree[rt * 2 + 1].b - tt;
  Tree[rt].ans = Tree[rt * 2].ans + Tree[rt * 2 + 1].ans + tt * 2;
}
void Build(int l, int r, int rt) {
  Tree[rt].l = l, Tree[rt].r = r;
  if (l == r) {
    if (s[l - 1] == '(')
      Tree[rt].a++;
    else
      Tree[rt].b++;
    Tree[rt].ans = 0;
    return;
  }
  int mid = (l + r) / 2;
  Build(l, mid, rt * 2);
  Build(mid + 1, r, rt * 2 + 1);
  pushup(rt);
}
int ans, now;
node queryans(int L, int R, int rt) {
  if (L == Tree[rt].l && Tree[rt].r == R) {
    return Tree[rt];
  }
  int mid = (Tree[rt].l + Tree[rt].r) / 2;
  if (R <= mid)
    return queryans(L, R, rt * 2);
  else if (L > mid)
    return queryans(L, R, rt * 2 + 1);
  else {
    node as;
    node as1 = queryans(L, mid, rt * 2);
    node as2 = queryans(mid + 1, R, rt * 2 + 1);
    int tt = min(as1.a, as2.b);
    as.a = as1.a + as2.a - tt;
    as.b = as1.b + as2.b - tt;
    as.ans = as1.ans + as2.ans + tt * 2;
    return as;
  }
}
int main() {
  cin >> s;
  int n = strlen(s);
  Build(1, n, 1);
  int q, l, r;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> l >> r;
    node tmp = queryans(l, r, 1);
    cout << tmp.ans << endl;
  }
}
