Eugeny and Array（思维）

佚名 4年前 (2020-01-14) 算法 956人围观抢沙发百度已收录

Eugeny has array a = a₁, a₂, ..., a_n, consisting of n integers. Each integer a_i equals to -1, or to 1. Also, he has m queries:

Query number i is given as a pair of integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n).
The response to the query will be integer 1, if the elements of array a can be rearranged so as the sum a_{l_i} + a_{l_i + 1} + ... + a_{r_i} = 0, otherwise the response to the query will be integer 0.

Help Eugeny, answer all his queries.

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Input

The first line contains integers n and m (1 ≤ n, m ≤ 2·10⁵). The second line contains n integers a₁, a₂, ..., a_n (a_i = -1, 1). Next m lines contain Eugene's queries. The i-th line contains integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n).

Output

Print m integers — the responses to Eugene's queries in the order they occur in the input.

Examples

Input

Output

0
1
0

Input

5 5
-1 1 1 1 -1
1 1
2 3
3 5
2 5
1 5

Output

0
1
0
1
0
题意：问这一组数重排后，能不能使区间[l，r]的和为0
这题就在卡输入
代码：

#include<iostream>
using namespace std;
int main(){
    ios::sync_with_stdio(false);//c++加速语句 
    int n,m;
    cin>>n>>m;
    int cnt1=0,cnt2=0;
    while(n--){
        int a;
        cin>>a;
        if(a>0)  cnt1++;
        else cnt2++;
    }
    int minn=min(cnt1,cnt2);
    while(m--){
        int l,r;
        cin>>l>>r;
        if((r-l+1)%2==0&&(r-l+1)/2<=minn) cout<<"1"<<endl;
        else cout<<"0"<<endl;
    }
    return 0;
}