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;
char s[1000100];
struct node {
  int f, s, b;
} Tree[4000100];
int q, a, b, n;
void build(int cn, int st, int ed) {
  if (st == ed) {
    Tree[cn].b = 0;
    if (s[st] == '(')
      Tree[cn].f = 1;
    else
      Tree[cn].s = 1;
    return;
  }
  int lft = cn << 1, rgt = lft + 1, mid = (st + ed) / 2;
  build(lft, st, mid);
  build(rgt, mid + 1, ed);
  int mn = min(Tree[lft].f, Tree[rgt].s);
  Tree[cn].b = Tree[lft].b + Tree[rgt].b + (mn << 1);
  Tree[cn].f = Tree[lft].f + Tree[rgt].f - mn;
  Tree[cn].s = Tree[lft].s + Tree[rgt].s - mn;
}
node query(int cn, int st, int ed, int L, int R) {
  if (st == L && ed == R) return Tree[cn];
  int lft = cn << 1, rgt = lft + 1, mid = (st + ed) / 2;
  if (R <= mid)
    return query(lft, st, mid, L, R);
  else if (L > mid)
    return query(rgt, mid + 1, ed, L, R);
  else {
    node tmpa = query(lft, st, mid, L, mid);
    node tmpb = query(rgt, mid + 1, ed, mid + 1, R);
    node tmp;
    int mn = min(tmpa.f, tmpb.s);
    tmp.b = tmpa.b + tmpb.b + (mn << 1);
    tmp.f = tmpa.f + tmpb.f - mn;
    tmp.s = tmpa.s + tmpb.s - mn;
    return tmp;
  }
}
int main() {
  memset(Tree, 0, sizeof Tree);
  scanf("%s", s + 1);
  n = strlen(s + 1);
  build(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    scanf("%d %d", &a, &b);
    printf("%d\n", query(1, 1, n, a, b).b);
  }
}
