G - Bear and Three Balls

最新推荐文章于 2024-06-23 01:44:59 发布

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本文介绍了一个有趣的问题：小熊Limak需要从多个球中挑选出三个球分别送给他的三位朋友，但选择球的大小需要遵循特定规则。文章通过示例说明了如何确保所选球的大小既不相同，又不超过规定范围内的差异。

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Limak is a little polar bear. He has n balls, the i-th ball has size t_i.

Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:

No two friends can get balls of the same size.
No two friends can get balls of sizes that differ by more than 2.

For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).

Your task is to check whether Limak can choose three balls that satisfy conditions above.

Input

The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.

The second line contains n integers t₁, t₂, ..., t_n (1 ≤ t_i ≤ 1000) where t_i denotes the size of the i-th ball.

Output

Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).

Example

Input

4
18 55 16 17

Output

YES

Input

6
40 41 43 44 44 44

Output

NO

Input

8
5 972 3 4 1 4 970 971

Output

YES

Note

In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.

In the second sample, there is no way to give gifts to three friends without breaking the rules.

In the third sample, there is even more than one way to choose balls:

Choose balls with sizes 3, 4 and 5.
Choose balls with sizes 972, 970, 971.

分析：3人所得球半径不能超过2；则a,b,c需满足a+1=b;a+2=c;（也就是相邻数据相差为1）

首先把重复数据删除，再排序和判断；

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
int a[1000],b[1000];
int main()
{int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
int pb=0,flag;
for(int i=0;i<n;i++)
{
flag=1;
for(int j=0;j<pb;j++)
if(b[j]==a[i])
flag=0;
if(flag)
b[pb++]=a[i];
}
int k=0;
sort(b,b+pb);
for(int i=0;i<pb;i++)
{
if(b[i]+1==b[i+1]&&b[i]+2==b[i+2])
k=1;
if(i+2>=pb-1)
break;
}
if(!k)
printf("NO\n");
else printf("YES");

return 0;
}