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 MIN = 1e3 + 1;
const int MXN = 1e6 + 1;
const int INF = 1e9 + 7;
const long long LINF = 1e18 + 15;
const double EPS = 1e-9;
string st;
int n, A, B, C;
struct brackets {
  int a, b, c;
} t[MXN * 4];
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v].a = 0;
    if (st[tl - 1] == '(')
      t[v].b = 1;
    else
      t[v].c = 1;
  } else {
    int tm = (tl + tr) / 2;
    build(v * 2, tl, tm);
    build(v * 2 + 1, tm + 1, tr);
    t[v].a = t[v * 2].a + t[v * 2 + 1].a + min(t[v * 2].b, t[v * 2 + 1].c) * 2;
    t[v].b = t[v * 2].b + t[v * 2 + 1].b - min(t[v * 2].b, t[v * 2 + 1].c);
    t[v].c = t[v * 2].c + t[v * 2 + 1].c - min(t[v * 2].b, t[v * 2 + 1].c);
  }
}
void get(int v, int tl, int tr, int l, int r) {
  if (l > r) {
    return;
  }
  if (l <= tl && tr <= r) {
    A = A + t[v].a + min(B, t[v].c) * 2;
    int BB = B;
    B = B + t[v].b - min(B, t[v].c);
    C = C + t[v].c - min(BB, t[v].c);
    return;
  }
  int tm = (tl + tr) / 2;
  get(v * 2, tl, tm, l, min(r, tm));
  get(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
}
int main() {
  cin >> st;
  build(1, 1, st.size());
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int l, r;
    cin >> l >> r;
    A = 0;
    B = 0;
    C = 0;
    get(1, 1, st.size(), l, r);
    cout << A << "\n";
  }
  return 0;
}
