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_{l_i}, s_{l_i + 1}, ..., s_{r_i}. 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

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+;

 const int maxv=1e5+;

 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=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

     return x*f;

 }

 inline void print(int x){

     if(x<)putchar('-'),x=-x;

     if(x>)print(x/);

     putchar(x%+'');

 }

 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=;

     for(int i=x;i>;i-=lowbit(i))ans+=tree[i];

     return ans;

 }

 int main(){

     while(~scanf("%s",str)){

         while(!st.empty())st.pop();n=strlen(str);

         memset(tree,,sizeof(tree)),memset(ans,,sizeof(ans));m=read();

         for(int i=;i<m;++i)query[i].l=read(),query[i].r=read(),query[i].id=i;

         sort(query,query+m,cmp);pos=;

         for(int i=;i<m;++i){

             for(int j=pos;j<=query[i].r;++j){

                 if(str[j-]=='(')st.push(j);///记录左括号的位置（物理位置）

                 else if(!st.empty())add(st.top(),),st.pop();///单点更新

             }

             pos=query[i].r+;///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-);

         }

         for(int i=;i<m;++i)print(ans[i]),puts("");

     }

     return ;

 }