IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2)

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 than2

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.

Inpu

The first line of the input contains one integer n (3 ≤ n ≤ 50) —
the number of balls Limak ha

The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000)
where ti denotes
the size of the i-th ball.

Outpu

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

题意：

A给3个朋友分苹果，无规矩不交友，规定每个朋友分的苹果大小不能相同，大小之差不可以超过2，给你n个苹果地大小，问能否选出3个苹果将它们分给朋友们；

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

<pre name="code" class="cpp"><pre name="code" class="cpp">#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
int n;
int d[60];
while(cin>>n)
{
for(int i=0;i<n;i++)
cin>>d[i];
sort(d,d+n);
int m=unique(d,d+n)-d;///消除相等的值
for(int i=2;i<m;i++)
{
if(d[i]==d[i-1]+1&&d[i]==d[i-2]+2)
{
cout<<"YES"<<endl;
goto endW;
}
}
cout<<"NO"<<endl;
endW:;
}
return 0;
}