Codeforces 653A： Bear and Three Balls（技巧，水）

A. Bear and Three Balls

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Limak is a little polar bear. He has n balls, the i-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 sizes 90, 91 and 92.
But he can't choose balls with sizes 5, 5and 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, ..., tn (1 ≤ ti ≤ 1000)
where ti 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).

Examples

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 sizes18, 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.

题目大意：给你一个序列，看是否有三个连续的整数。
解题思路：用数组记录某数是否存在：刚开始将数组清零，然后在输入时，输入x，则让a[x]=1即可，最后将数组扫一遍，看是否存在连续的三个整数。

代码如下：

#include <stdio.h>
#include <string.h>
int main()
{
int n;
int flag[1010];
memset(flag,0,sizeof(flag));
int x;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&x);
flag[x]=1;
}
for(int i=0;i<1010;i++)
{
if(flag[i]==1&&flag[i+1]==1&&flag[i+2]==1)
{
printf("YES\n");
return 0;
}
}
printf("NO\n");
return 0;
}