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

• Query number?i?is given as a pair of integers?li,?ri?(1?≤?li?≤?ri?≤?n).
• The response to the query will be integer?1, if the elements of array?a?can be rearranged so as the sum?ali?+?ali?+?1?+?...?+?ari?=?0, otherwise the response to the query will be integer?0.

Help Eugeny, answer all his queries.

Input

The first line contains integers?n?and?m?(1?≤?n,?m?≤?2·105). The second line contains?n?integers?a1,?a2,?...,?an?(ai?=?-1,?1). Next?m?lines contain Eugene's queries. The?i-th line contains integers?li,?ri?(1?≤?li?≤?ri?≤?n).

Output

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

Examples

Input

2 3 1 -1 1 1 1 2 2 2

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

代碼

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm>using namespace std;int main() {int n,m;int num[200005];cin>>n>>m;int sum1=0,sum2=0;for(int t=0; t<n; t++) {scanf("%d",&num[t]);if(num[t]==-1) {sum1++;} else {sum2++;}}int l,r;for(int t=0; t<m; t++) {scanf("%d%d",&l,&r);if((r-l)%2==0)printf("0\n");else if((r-l)%2!=0) {if((r-l+1)/2<=sum1&&(r-l+1)/2<=sum2)printf("1\n");elseprintf("0\n");}}return 0; }

?

轉(zhuǎn)載于:https://www.cnblogs.com/Staceyacm/p/10781942.html

總結(jié)

以上是生活随笔為你收集整理的Eugeny and Array（水题，注意题目描述即可）的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

如果覺(jué)得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。