Codeforces Round #212 (Div. 2) B. Petya and Staircases

B. Petya and Staircases

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 numbern 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 andm (1 ≤ n ≤ 109,0 ≤ m ≤ 3000)
— the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line containsm
different space-separated integersd1, d2, ..., dm (1 ≤ di ≤ n)
— the numbers of the dirty stairs (in an arbitrary order).

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

Examples

Input

10 5
2 4 8 3 6

Output

NO

Input

10 5
2 4 5 7 9

Output

YES

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
int n,i,m,a[3005];
memset(a,0,sizeof(a));
scanf("%d%d",&n,&m);
for(i=0;i<m;i++)
cin >> a[i];
sort(a,a+m);
for(i=2;i<m;i++)
{
if(a[i]-a[i-1]==1&&a[i-1]-a[i-2]==1)
{
printf("NO\n");
return 0;
}
}
if(a[0]!=1&&a[m-1]!=n)
printf("YES\n");
else
printf("NO\n");
return 0;
}