Sereja and Brackets(括号匹配)

Description

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 l_i, r_i(1 ≤ l_i ≤ r_i ≤ n). The answer to the i-th query is the length of the maximum correct bracket subsequence of sequence s_li, s_li + 1, ..., s_ri. 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 l_i, r_i (1 ≤ l_i ≤ r_i ≤ 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.

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 = s_k1s_k2... 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 «()(())(())».

解题思路：求连续子串中括号匹配的个数。括号匹配问题一般可以用栈来维护。对于每个右括号，其匹配的左括号是固定的，因此我们可以记录每个右括号匹配的左括号位置，对整个区间进行线扫描，同时用树状数组维护+离散化处理即可。

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e6+5;
const int maxv=1e5+5;
char str[maxn];int n,m,pos,tree[maxn],ans[maxv];stack<int> st;
struct node{int l,r,id;}query[maxn];
bool cmp(node a,node b){return a.r<b.r;}///右端点排序
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void print(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)print(x/10);
    putchar(x%10+'0');
}
int lowbit(int x){return x&-x;}
void add(int x,int val){
    for(int i=x;i<=n;i+=lowbit(i))tree[i]+=val;
}
int get_sum(int x){
    int ans=0;
    for(int i=x;i>0;i-=lowbit(i))ans+=tree[i];
    return ans;
}
int main(){
    while(~scanf("%s",str)){
        while(!st.empty())st.pop();n=strlen(str);
        memset(tree,0,sizeof(tree)),memset(ans,0,sizeof(ans));m=read();
        for(int i=0;i<m;++i)query[i].l=read(),query[i].r=read(),query[i].id=i;
        sort(query,query+m,cmp);pos=1;
        for(int i=0;i<m;++i){
            for(int j=pos;j<=query[i].r;++j){
                if(str[j-1]=='(')st.push(j);///记录左括号的位置（物理位置）
                else if(!st.empty())add(st.top(),2),st.pop();///单点更新
            }
            pos=query[i].r+1;///从下一个位置开始，避免重叠，O(n)遍历
            ans[query[i].id]=get_sum(query[i].r)-get_sum(query[i].l-1);
        }
        for(int i=0;i<m;++i)print(ans[i]),puts("");
    }
    return 0;
}