codeforces-653

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A. Bear and Three Balls

time limit per test

memory limit per test

input

output

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.
• 2.

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

n (3 ≤ n ≤ 50) — the number of balls Limak has.

n integers t1, t2, ..., tn (1 ≤ ti) where ti denotes the size of the i-th ball.

Output

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

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:

1. 3, 4 and 5.
2. 972, 970, 971.

题意：一个人要送球给他三个朋友，每个朋友的球大小不能一样，而且任意两个朋友的球大小不能超过 2；

傻逼还去重：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
int a[100];
int main()
{
  while(~scanf("%d",&n))
  {
    for(int i=0;i<n;i++)
      scanf("%d",&a[i]);
    sort(a,a+n);
    bool flag=0;
    int top=unique(a,a+n)-a; // 去重 
    if(top<3)
    {
      puts("NO");
      continue;
    }
    for(int i=1;i<top-1;i++)
    {
      if(a[i]==a[i-1]+1&&a[i+1]==a[i]+1)
      {
        flag=1;
        break;
      }
    }
    if(flag)  puts("YES");
    else  puts("NO");
  }
  return 0;
}

n的范围也太小了吧，三重 for 水

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
int a[100];
int main()
{
  while(~scanf("%d",&n))
  {
    for(int i=0;i<n;i++)
      scanf("%d",&a[i]);
    bool flag=0;
    for(int i=0;i<n;i++)
    {
      for(int j=0;j<n;j++)
      {
        for(int k=0;k<n;k++)
        {
          if(a[i]==a[j]+1&&a[i]==a[k]-1)
            flag=1;
        }
      }
    }
    if(flag)  puts("YES");
    else  puts("NO");
  }
  return 0;
}