#include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
  int n,k;
  while(cin>>n>>k,n){
    int x[k];
    for(int i=0;i<k;i++) cin>>x[i];
    int e[n+2];
    memset(e,0,sizeof(e));
    for(int i=0;i<k;i++) e[x[i]]=1;
    int c[2][n+2];
    memset(c,0,sizeof(c));
    for(int i=1;i<=n;i++)
      if(e[i]) c[0][i]=c[0][i-1]+1;
    for(int i=n;i>=0;i--)
      if(e[i]) c[1][i]=c[1][i+1]+1;
    int ans=0;
    for(int i=1;i<=n;i++){
      if(e[0]&&!e[i])
	ans=max(ans,c[0][i-1]+c[1][i+1]+1);
      ans=max(ans,c[0][i]);
      ans=max(ans,c[1][i]);
    }
    cout<<ans<<endl;
  }
  return 0;
}
### Prompt

Your task is to create a Cpp solution to the following problem:

problem

There are n cards with one integer from 1 to n and one blank card. Of these n + 1 cards, k cards are given, but 1 ≤ k ≤ n. You can write one integer from 1 to n on a blank card. I want to make a continuous sequence of integers as long as possible with just a given card.

Write a program that outputs the maximum length of a contiguous sequence of integers that can be made from a given card when the given card is entered.



input

The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing two zeros.

On the first line, two integers n (1 ≤ n ≤ 100000) and k (1 ≤ k ≤ n) are written in this order, separated by one blank. The following k line contains one integer. Written one by one, representing the integers written on the given k cards. Blank cards are represented by 0.

Of the scoring data, 40% of the points are 1 ≤ n ≤ 1000, 1 ≤ k ≤ 500, 20% of the points are 1 ≤ n ≤ 60000, 1 ≤ k ≤ 50000. Satisfy 1 ≤ n ≤ 100000, 1 ≤ k ≤ 100000.

The number of datasets does not exceed 5.

output

Outputs an integer on one line for each dataset.

Examples

Input

7 5
6
2
4
7
1
7 5
6
2
0
4
7
0 0


Output

2
4


Input

None


Output

None

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
  int n,k;
  while(cin>>n>>k,n){
    int x[k];
    for(int i=0;i<k;i++) cin>>x[i];
    int e[n+2];
    memset(e,0,sizeof(e));
    for(int i=0;i<k;i++) e[x[i]]=1;
    int c[2][n+2];
    memset(c,0,sizeof(c));
    for(int i=1;i<=n;i++)
      if(e[i]) c[0][i]=c[0][i-1]+1;
    for(int i=n;i>=0;i--)
      if(e[i]) c[1][i]=c[1][i+1]+1;
    int ans=0;
    for(int i=1;i<=n;i++){
      if(e[0]&&!e[i])
	ans=max(ans,c[0][i-1]+c[1][i+1]+1);
      ans=max(ans,c[0][i]);
      ans=max(ans,c[1][i]);
    }
    cout<<ans<<endl;
  }
  return 0;
}
```