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 = (int)1e5 + 17;
const int MX = (int)1e6 + 17;
const int MOD = (int)1e9 + 7;
const long long oo = LLONG_MAX;
const int INF = INT_MAX;
const long double Pi = 3.14159265358979323846264338327950288419716939937510;
const int di[4] = {-1, 0, 1, 0};
const int dj[4] = {0, 1, 0, -1};
inline long long IN() {
  long long x = 0, ch = getchar(), f = 1;
  while (!isdigit(ch) && (ch != '-') && (ch != EOF)) ch = getchar();
  if (ch == '-') {
    f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void OUT(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10)
    OUT(x / 10), putchar(x % 10 + '0');
  else
    putchar(x + '0');
}
void IOI2017() {}
int n, m, l, r;
char a[MX];
struct T {
  int l, r, s;
} t[MX * 4];
inline T Merge(T x, T y) {
  int add = min(x.l, y.r);
  T ans;
  ans.s = x.s + y.s + add;
  ans.l = x.l + y.l - add;
  ans.r = x.r + y.r - add;
  return ans;
}
inline void build(int v = 1, int tl = 1, int tr = n) {
  if (tl == tr) {
    if (a[tl] == '(')
      t[v].l++;
    else
      t[v].r++;
  } else {
    int tm = (tl + tr) >> 1;
    build(v << 1, tl, tm);
    build((v << 1) + 1, tm + 1, tr);
    t[v] = Merge(t[v << 1], t[(v << 1) + 1]);
  }
}
inline T get(int l, int r, int v = 1, int tl = 1, int tr = n) {
  if (l <= tl && tr <= r) return t[v];
  if (tl > r || tr < l)
    return {0, 0, 0};
  else {
    int tm = tl + tr >> 1;
    return Merge(get(l, r, v << 1, tl, tm),
                 get(l, r, (v << 1) + 1, tm + 1, tr));
  }
}
int main() {
  gets(a);
  n = (int)strlen(a);
  for (int i = n; i > 0; i--) a[i] = a[i - 1];
  build();
  m = IN();
  while (m--) {
    l = IN(), r = IN();
    OUT(get(l, r).s << 1), putchar('\n');
  }
  return 0;
}
