CF w4d1 B. Petya and Staircases

Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them.

Now Petya is on the first stair of the staircase, consisting of n stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number n without touching a dirty stair once.

One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only.

Input

The first line contains two integers n and m (1 ≤ n ≤ 10^9, 0 ≤ m ≤ 3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains m different space-separated integers d 1, d 2, ..., d m (1 ≤ d i ≤ n) — the numbers of the dirty stairs (in an arbitrary order).

Output

Print "YES" if Petya can reach stair number n, stepping only on the clean stairs. Otherwise print "NO".

Examples

inputCopy
10 5
2 4 8 3 6
outputCopy
NO

inputCopy
10 5
2 4 5 7 9
outputCopy
YES

小男孩一次可以跳一格或者两格或者三格，要是有三个连续的脏楼梯就过不去了。有三个特判：第一个楼梯是脏的或者最后一个楼梯是脏的或者没有脏楼梯。

#include<bits/stdc++.h>
using namespace std;
int f[3005];
int n,m;
bool ans=true;

int main()
{
	cin>>n>>m;
	for(int i=0;i<m;i++)cin>>f[i];
	sort(f,f+m);
	if(f[0]==1||f[m-1]==n||m==0){//特判
		if(m==0)cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
		return 0;
	}
	for(int i=0;i<m-2;i++){
		if(f[i]=f[i+1]-1&&f[i]==f[i+2]-2){//连续3个脏楼梯
			ans=false;
			break;
		}
	}
	if(ans)cout<<"YES"<<endl;
	else cout<<"NO"<<endl;
	return 0;
}