Fashion in Berland

最新推荐文章于 2023-01-09 13:39:23 发布

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本文介绍了一个简单的程序，用于根据Berland时尚规则判断一件带有n个纽扣的夹克是否正确扣好。根据规则，除了一个纽扣外，所有纽扣都应被扣上，且未扣上的纽扣不必是最后一个。当夹克只有一个纽扣时，该纽扣必须扣上。

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Description

According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.

You are given a jacket with n buttons. Determine if it is fastened in a right way.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of buttons on the jacket.

The second line contains n integers a_i (0 ≤ a_i ≤ 1). The number a_i = 0 if the i-th button is not fastened. Otherwise a_i = 1.

Output

In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".

Sample Input

Input

3
1 0 1

Output

YES

Input

3
1 0 0

Output

NO

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

int a[1000+10];

int main()
{
	int num;
	int i,k;
	scanf("%d",&num);
	int n=0;
	for(i=0;i<num;i++)
	{
		scanf("%d",&a[i]);
		if(a[i]==0)
			n++;
	}
	if(num==1)
	{
		if(a[0]==1)
			k=1;
		else
			k=0;
	}
	else
	{
		if(n==1)
			k=1;
		else
			k=0;
	}
	if(k)
		printf("YES\n");
	else
		printf("NO\n");
	return 0;
}