CodeForces - 362B 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 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, ..., dm (1 ≤ di ≤ 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

input

10 5
2 4 8 3 6

output

NO

input

10 5
2 4 5 7 9

output

YES

数出最多连续的凳子数，注意脏的凳子数可能为0

#include <iostream>
#include <cstdio>
#include <algorithm>
#define max_ 3010
using namespace std;
int n,m;
int num[max_];
int main()
{
cin>>n>>m;
for(int i=1;i<=m;i++)
{
cin>>num[i];
}
if(m==0)
{
printf("YES\n");
return 0;
}
sort(num+1,num+1+m);
if(num[1]==1||num[m]==n)
{
printf("NO\n");
return 0;
}
int i,cnt=1;
for(i=2;i<=m;i++)
{
if(num[i]-num[i-1]==1)
{
cnt++;
if(cnt>2)
break;
}
else
{
cnt=1;
}
}
if(i==m+1)
printf("YES\n");
else
printf("NO\n");
}