Bear and Three Balls

                                 Bear and Three Balls

（sort）Time Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

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Description

Limak is a little polar bear. He has n balls, the i-th ball has sizet_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 than2.

For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes90, 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 sizes30, 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 integerst₁, t₂, ..., t_n (1 ≤ t_i ≤ 1000) where t_i denotes the size of thei-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 than2. Otherwise, print "NO" (without quotes).

Sample Input

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判断是否存在三个连续自然数，
  #include<stdio.h>      int num[1001];      int main()      {          int n,t;          scanf("%d",&n);          while(n--)          {              scanf("%d",&t);              num[t]++;          }          for(int i=0;i<=1000;i++)          {              if(num[i]&&num[i+1]&&num[i+2])              {                  printf("YES\n");                  return 0;              }          }          printf("NO\n");          return 0;         }  

另附错误代码一份，求改
#include<stdio.h>#include<algorithm>#include<stdlib.h>#include<iostream>using namespace std;int main(){int n,a[1010];while(~scanf("%d",&n)){memset(a,0,sizeof(a));int k=0;for(int i=0;i<n;i++){scanf("%d",&a[i]);}sort(a,a+n);for(int i=1;i<n;i++){if(a[i]==a[i-1])for(int j=i;j<n-1;j++)a[j]=a[j+1];}for(int i=1;i<n-1;i++){if(a[i-1]+1==a[i]&&a[i]+1==a[i+1]){k=1;break;}}if(k==1)printf("YES\n");elseprintf("NO\n");}}