#include <bits/stdc++.h>
using namespace std;

int n, k;
int cnt;
bool f[101010];

int main()
{
  while (cin >> n >> k, n){
    memset(f, false, sizeof(f));
    cnt = 0;

    for (int i = 0; i < k; i++){
      int x;
      cin >> x;
      if (x--) f[x] = true;
      else cnt++;
    }

    int max_d = -1;
    queue<int> q;

    int l = 0, r = 0;
    while (q.size() <= cnt){
      while (f[r]){
        r++;
        if (r + 1 == n) break;
      }
      q.push(r);

      max_d = max(max_d, r - l);
      r++;
      if (r >= n) break;
      if (q.size() == cnt + 1){
        l = q.front(); q.pop();
      }
    }

    cout << max_d - 1 << endl;
  }
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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;

int n, k;
int cnt;
bool f[101010];

int main()
{
  while (cin >> n >> k, n){
    memset(f, false, sizeof(f));
    cnt = 0;

    for (int i = 0; i < k; i++){
      int x;
      cin >> x;
      if (x--) f[x] = true;
      else cnt++;
    }

    int max_d = -1;
    queue<int> q;

    int l = 0, r = 0;
    while (q.size() <= cnt){
      while (f[r]){
        r++;
        if (r + 1 == n) break;
      }
      q.push(r);

      max_d = max(max_d, r - l);
      r++;
      if (r >= n) break;
      if (q.size() == cnt + 1){
        l = q.front(); q.pop();
      }
    }

    cout << max_d - 1 << endl;
  }
}
```