暑期集训之Bear and Three Balls

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

418 55 16 17

Output

YES

Input

640 41 43 44 44 44

Output

NO

Input

85 972 3 4 1 4 970 971

Output

YES

这道题虽然说是sort排序专题的题目，但却是也是一道小思维题，说了一大堆，说白了就是看选的球里面有没有三个连续大小的球，既然我们知道这个了，其实就很简单了，不过，虽然看上去挺简单的，但依旧用到了一点小技巧，详解在代码里面，代码如下：

#include<stdio.h>#include<string.h>int main(){     int n,flag=0;    int k[1005];//这个是放球的尺寸的数组     int m[1005];//这个是标记数组     memset(m,0,sizeof(m)); //归零操作现在很常见，所以要多加熟练，这里归零的话是为了方便下面的标记操作     scanf("%d",&n);    for(int i=0;i<n;i++)    {    scanf("%d",&k[i]);    m[k[i]]=1;//这里就是标记操作了，把拥有的尺寸的球都给标记出来 }for(int i=1;i<=1000;i++)//这个循环其实是一个查找的过程，从尺寸1开始到1000，看有没有三个连续尺寸的球被我们标记出来了 {if(m[i]==1&&m[i+1]==1&&m[i+2]==1)flag=1;//flag标记也很常用了，尤其在循环里面用的很多 }if(flag==1)printf("Yes\n");elseprintf("No\n");return 0;}//想通了这道题也还是非常简单的ORZ 

今天的专题训练最后还有一个简单贪心的题，但无奈都不知道贪心是啥QAQ，看了看别人的博客也一时半会没搞懂，但会努力学习的ORZ，今天就到此结束~\(≧▽≦)/~啦啦啦