CF380C. Sereja and Brackets[线段树 区间合并]

C. Sereja and Brackets

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 «()(())(())».

---

一定要读懂题

问的是匹配的括号子序列的最大长度

其实也就是最多有几对匹配*2

---

用线段树做

We will support the segments tree. At each vertex will be stored:
av — the maximum length of the bracket subsequence
bv — how many there it open brackets that sequence doesn't contain
cv — how many there it closed brackets that sequence doesn't contain
If we want to combine two vertices with parameters (a1, b1, c1) and (a2, b2, c2), we can use the following rules:
t = min(b1, c2)
a = a1 + a2 + t
b = b1 + b2 - t
c = c1 + c2 - t

读懂题就很明白了，就是个合并问题

查询的时候可以像GSS1那样写qpre和qsuf，但这次采用一个新方法，merge和query返回节点，这样方便好多

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define m ((l+r)>>1)
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define lc o<<1
#define rc o<<1|1
using namespace std;
typedef long long ll;
const int N=1e6+5,INF=2e9+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
char s[N];
int q,ql,qr;
struct node{
    int a,b,c;
    node():a(0),b(0),c(0){}
}t[N<<2];

inline node merge(node x,node y){
    node z;
    int t=min(x.b,y.c);
    z.a=x.a+y.a+t;
    z.b=x.b+y.b-t;
    z.c=x.c+y.c-t;
    //printf("merge %d %d %d
",z.a,z.b,z.c);
    return z;
}
void build(int o,int l,int r){
    if(l==r){
        if(s[l]=='(') t[o].b=1;
        else t[o].c=1;
    }else{
        build(lson);
        build(rson);
        t[o]=merge(t[lc],t[rc]);
    }
}
node query(int o,int l,int r,int ql,int qr){
    if(ql<=l&&r<=qr) return t[o];
    else{
        node ans;
        if(ql<=m) ans=merge(ans,query(lson,ql,qr));
        if(m<qr) ans=merge(ans,query(rson,ql,qr));
        return ans;
    }
}
int main(){
    scanf("%s",s+1);
    q=read();
    int n=strlen(s+1);
    build(1,1,n);
    for(int i=1;i<=q;i++){
        ql=read();qr=read();
        printf("%d
",query(1,1,n,ql,qr).a*2);
    }
}