#include <iostream>
#include <algorithm>

#define rep(i, n) for(int i = 0; i < (n); i++)

using namespace std;

int v[100005];
int main(){
    int n, k;
    while(cin >> n >> k, n|k){
        rep(i, k) cin >> v[i];
		sort(v, v+k);
        int prev = 0, seq = 0;
        int ans = 0;
        for(int i = 1+!v[0]; i <= k; i++){
            seq++;
            if(v[i] != v[i-1]+1 || i == k){
				if(!v[0]) ans = max(ans, prev+seq+1);
				else ans = max(ans, seq);
                prev = seq;
                seq = 0;
            }
        }
        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 <iostream>
#include <algorithm>

#define rep(i, n) for(int i = 0; i < (n); i++)

using namespace std;

int v[100005];
int main(){
    int n, k;
    while(cin >> n >> k, n|k){
        rep(i, k) cin >> v[i];
		sort(v, v+k);
        int prev = 0, seq = 0;
        int ans = 0;
        for(int i = 1+!v[0]; i <= k; i++){
            seq++;
            if(v[i] != v[i-1]+1 || i == k){
				if(!v[0]) ans = max(ans, prev+seq+1);
				else ans = max(ans, seq);
                prev = seq;
                seq = 0;
            }
        }
        cout << ans << endl;
    }
	return 0;
}
```