...(Div. 1 + Div. 2) A. Bear and Three Balls（是否存在三个连续自然数）

Limak is a little polar bear. He has n balls, thei-th ball has size
ti.

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 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 integers
t1, t2, ..., tn (1 ≤ ti ≤ 1000) whereti
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

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题解
给出一组数据，问是否能从中找出三个连续的数
先快排，再计数：
代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int n,a[10000];
while(~scanf("%d",&n))
{int i,j,k=1;
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]+1)
k++;
if((a[i]!=a[i-1]+1)&&a[i]!=a[i-1])//防止有重复数据
k=1;
if(k==3)//当有三个时输出YES{
printf("YES\n");
break;
}
}
if(k!=3)
printf("NO\n");
}
}