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 N = 1e6 + 10;
const int INF = 1e8;
int tree[N << 2], lazy[N << 2];
void go(int s, int e, int ind) {
  tree[ind] += lazy[ind];
  if (s != e) {
    lazy[ind << 1] += lazy[ind];
    lazy[ind << 1 | 1] += lazy[ind];
  }
  lazy[ind] = 0;
}
int n;
void update(int l, int r, int add, int s = 1, int e = n, int ind = 1) {
  go(s, e, ind);
  if (l > e || s > r) return;
  if (s >= l && e <= r) {
    lazy[ind] += add;
    go(s, e, ind);
    return;
  }
  int mid = (s + e) >> 1;
  update(l, r, add, s, mid, ind << 1);
  update(l, r, add, mid + 1, e, ind << 1 | 1);
  tree[ind] = min(tree[ind << 1], tree[ind << 1 | 1]);
}
int get(int l, int r, int s = 1, int e = n, int ind = 1) {
  go(s, e, ind);
  if (s >= l && e <= r) return tree[ind];
  if (s > r || l > e) return INF;
  int mid = (s + e) >> 1;
  return min(get(l, r, s, mid, ind << 1), get(l, r, mid + 1, e, ind << 1 | 1));
}
vector<pair<int, int>> queries[N];
char s[N];
int l[N], r[N], ans[N], pref[N];
int main() {
  scanf("%s", s);
  n = strlen(s);
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", l + i, r + i);
    queries[r[i]].push_back({l[i], i});
  }
  for (int i = 1; i <= n; i++) {
    char c = s[i - 1];
    pref[i] = pref[i - 1];
    if (c == '(') {
      pref[i]++;
      update(i, i, pref[i - 1]);
    } else {
      update(1, i, 1);
      update(i, i, INF);
    }
    for (auto it : queries[i]) {
      int val = get(it.first, i) - pref[it.first - 1];
      if (val > n) val = 0;
      val = min(val, pref[i] - pref[it.first - 1]);
      ans[it.second] = 2 * val;
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
