Sereja and Brackets

题目链接： CodeForces - 380C

Sereja has a bracket sequence s1, s2, ..., *s**n, 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, ..., *s**n* (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

())(())(())(71 12 31 21 128 125 112 10

Output

00210466

Note

A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is string x = sk1sk2... *s**k|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 «()(())(())».

题意：

给你一个只含有'(' 和')' 的字符串，

以及q个询问，每一个询问给你两个整数l和r，代表一个区间。对于每一个询问，让你输出区间中能选出最长的子序列是合法的括号序列的长度。

思路：

线段树+分治的思想来解决此问题。

我们线段树每一个区间维护以下信息：

1、区间中能选出最长的子序列是合法的括号序列的个数 num。

2、 区间中多余的'(' 字符的个数 a

3、区间中多余的')' 字符的个数 b

那么对于区间合并时，

num=左儿子的num+右儿子的num+min（左儿子的a，右儿子的b）

a=左儿子的a+右儿子的a - min（左儿子的a，右儿子的b）

b=左儿子的b+右儿子的b - min（左儿子的a，右儿子的b）

最后输出时，注意num个括号个数，*2才是长度。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2) { ans = ans * a % MOD; } a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
struct node {int l, r;int num;int a;// (int b;// )
} segmeng_tree[maxn << 2];
char s[maxn];
int n;
int m;
void pushup(int rt)
{int x = min(segmeng_tree[rt << 1].a, segmeng_tree[rt << 1 | 1].b);segmeng_tree[rt].num = x + segmeng_tree[rt << 1].num + segmeng_tree[rt << 1 | 1].num;segmeng_tree[rt].a = segmeng_tree[rt << 1].a + segmeng_tree[rt << 1 | 1].a - x;segmeng_tree[rt].b = segmeng_tree[rt << 1].b + segmeng_tree[rt << 1 | 1].b - x;
}
void build(int rt, int l, int r)
{segmeng_tree[rt].l = l;segmeng_tree[rt].r = r;if (l == r) {segmeng_tree[rt].a = s[l] == '(';segmeng_tree[rt].b = s[l] == ')';segmeng_tree[rt].num = 0;} else {int mid = (l + r) >> 1;build(rt << 1, l, mid);build(rt << 1 | 1, mid + 1, r);pushup(rt);}
}node ask(int rt, int l, int r)
{if (segmeng_tree[rt].l >= l && segmeng_tree[rt].r <= r) {return segmeng_tree[rt];}int mid = (segmeng_tree[rt].l + segmeng_tree[rt].r) >> 1;if (r <= mid) {return ask(rt << 1, l, r);} else if (l > mid) {return ask(rt << 1 | 1, l, r);} else {node res1 = ask(rt << 1, l, r);node res2 = ask(rt << 1 | 1, l, r);node res = res1;int x = min(res1.a, res2.b);res.num += x;res.b += res2.b;res.a += res2.a;res.num += res2.num;res.b -= x;res.a -= x;return res;}
}
int main()
{//freopen("D:\\code\\text\\input.txt","r",stdin);//freopen("D:\\code\\text\\output.txt","w",stdout);scanf("%s", s + 1);n = strlen(s + 1);build(1, 1, n);scanf("%d", &m);while (m--) {int l, r;scanf("%d %d", &l, &r);printf("%d\n", ask(1, l, r).num * 2);}return 0;
}inline void getInt(int *p)
{char ch;do {ch = getchar();} while (ch == ' ' || ch == '\n');if (ch == '-') {*p = -(getchar() - '0');while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 - ch + '0';}} else {*p = ch - '0';while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 + ch - '0';}}
}