#include <bits/stdc++.h>
#define FOR(i,n) for(int i=0;i<(int)(n);i++)
#define FORR(i,m,n) for(int i=(int)(m);i<(int)(n);i++)
#define pb(a) push_back(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define len(a) sizeof(a)
#define ll long long
#define INF 1<<29
using namespace std;

int a[101010];

void solve(){
 int n,k;
 for(;;){
  scanf("%d %d",&n,&k);
  if(n==0) break;
  FOR(i,k) scanf("%d",&a[i]);
  sort(a,a+k);
  int ma=0;
  FOR(i,k){
   if(a[i]==0) continue;
   FORR(j,i+1,k){
    if(!a[0]){
     if(a[j]-a[i]<=j-i+1) ma=max(ma,a[j]-a[i]+1);
     else break;
    }
    else{
     if(a[j]-a[i]==j-i) ma=max(ma,a[j]-a[i]+1);
     else break;
    }
   }
  }
  printf("%d\n",ma);
 }
}

int main(){
 solve();
 return 0;
}
### Prompt

Create a solution in cpp for 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>
#define FOR(i,n) for(int i=0;i<(int)(n);i++)
#define FORR(i,m,n) for(int i=(int)(m);i<(int)(n);i++)
#define pb(a) push_back(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define len(a) sizeof(a)
#define ll long long
#define INF 1<<29
using namespace std;

int a[101010];

void solve(){
 int n,k;
 for(;;){
  scanf("%d %d",&n,&k);
  if(n==0) break;
  FOR(i,k) scanf("%d",&a[i]);
  sort(a,a+k);
  int ma=0;
  FOR(i,k){
   if(a[i]==0) continue;
   FORR(j,i+1,k){
    if(!a[0]){
     if(a[j]-a[i]<=j-i+1) ma=max(ma,a[j]-a[i]+1);
     else break;
    }
    else{
     if(a[j]-a[i]==j-i) ma=max(ma,a[j]-a[i]+1);
     else break;
    }
   }
  }
  printf("%d\n",ma);
 }
}

int main(){
 solve();
 return 0;
}
```