Bear and Three Balls_综合

点击打开链接

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

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.

题解：这种纯英文的题对于我这种英语很渣的人来说是很吃力很吃力的，这题也是听了同学的翻译才知道这是一个水题，但我的第一个思路是错的，后面我又重新换了一种思路，第一种思路就是这样”a[i]+1==a[i+1]......“以此类推，后来我才发现这种方法是不对的，压根不行，我后面的思路就是把原来的数组放在一个新数组里面，让他同时等于一个数，然后如果新数组如果连续三个相等，然后再k++，最后输出k就可以了。

下面是我的思路：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstring>
using namespace std;
int a[10000],b[10000];
bool cmp(int a,int b)
{return a<b;
}
int  main()
{int n;while(scanf("%d",&n)!=EOF){int k=0;for(int i=0;i<n;i++){scanf("%d",&a[i]);b[a[i]]=1;}for(int i=0;i<9000;i++){if(b[i]==1&&b[i+1]==1&&b[i+2]==1){k=1;break;}}if(k==1){printf("YES\n");}elseprintf("NO\n");}return 0;
}