Codeforce Round #212 Div2 B

B. Petya and Staircases

time limit per test 1 second

memory limit per test 256 megabytes

 

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 ≤ 10⁹, 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 d₁, d₂, ..., 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".

Sample test(s)

Input

10 5 
2 4 8 3 6

Output

NO

Input

10 5 
2 4 5 7 9

Output

YES

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 30005
int a[maxn];
int main(){
    int t, n, m;
    scanf("%d%d",&n,&m);
    for (int i = 0; i < m; i++){
        scanf("%d", &a[i]);
    }
    sort(a, a + m);
    int flag = 1;
    if (a[0] == 1 || a[m - 1] == n){ printf("NO\n"); return 0; }
    for (int i = 0; i < m-2; i++){
        if (a[i] + 1 == a[i + 1] && a[i] + 2 == a[i + 2]){ flag = 0; break; }
    }
    if (flag){ printf("YES\n"); }
    else printf("NO\n");
    return 0;
}