【CodeForces 635A】 Bear and Three Balls

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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, thei-th ball has sizeti.

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 and3, or balls with sizes90,91 and
92. But he can't choose balls with sizes5,
5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes30,31 and
33 (because sizes30 and
33 differ by more than2).

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).

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.

//求所给数中有没有连续的三位数。

//代码如下：

#include <stdio.h>
#include <algorithm>
using namespace std;
int main()
{
int n,m,k,i;
int a[7777];
while(scanf("%d",&n)!=EOF)
{
k=0;
for(i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
for(i=0;i<n-1;)
{
if(a[i+1]-a[i]==1)
{
k++;
i=i+1;
if(k>=2)
{
printf("YES\n");
break;
}
}
else if(a[i+1]-a[i]==0)
{
i=i+1;
}
else
{
k=0;
i=i+1;
}
}
if(k<2)
printf("NO\n");
}
return 0;
}