#include <bits/stdc++.h>
#define r(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef pair<int,int>P;
int n,k,a[100001];
int main(){
	while(cin>>n>>k,n){
		vector<P>v;
		int ans=0,p=0;
		r(i,k){
			cin>>a[i];
			if(!a[i]){
				i--;
				k--;
				p++;
			}
		}
		sort(a,a+k);
		r(i,k){
			int res=1;
			while(i!=k-1&&a[i]==a[i+1]-1){
				res++;i++;
			}
			ans=max(ans,res);
			v.push_back(P(1,res));
			v.push_back(P(0,a[i+1]-a[i]));
		}
		if(p)r(i,v.size()){
			if(v[i].first==0&&v[i].second<=2){
				ans=max(ans,v[i-1].second+v[i+1].second+1);
			}
		}
		cout<<ans<<endl;
	}
}
### Prompt

Please formulate 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>
#define r(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef pair<int,int>P;
int n,k,a[100001];
int main(){
	while(cin>>n>>k,n){
		vector<P>v;
		int ans=0,p=0;
		r(i,k){
			cin>>a[i];
			if(!a[i]){
				i--;
				k--;
				p++;
			}
		}
		sort(a,a+k);
		r(i,k){
			int res=1;
			while(i!=k-1&&a[i]==a[i+1]-1){
				res++;i++;
			}
			ans=max(ans,res);
			v.push_back(P(1,res));
			v.push_back(P(0,a[i+1]-a[i]));
		}
		if(p)r(i,v.size()){
			if(v[i].first==0&&v[i].second<=2){
				ans=max(ans,v[i-1].second+v[i+1].second+1);
			}
		}
		cout<<ans<<endl;
	}
}
```