#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <algorithm>

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

using namespace std;

int main() {
  while(1) {
    int n,k;
    cin >> n >> k;
    if (!n) break;
    vector<int> v(n+1);
    for (int i = 0; i < k; ++i) {
      int a;
      cin >> a;
      v[a] = 1;
    }
    int l=0;
    int zero = 0;
    int max_len = 0;
    for (int i = 1; i <= n; ++i) {
      if (v[i] == 0) {
        zero++;
      }
      while(zero > v[0]) {
        if(v[++l] == 0) {
          zero--;
        }
      }
      max_len = max(max_len, i - l);
    }
    cout << max_len << endl;
  }
  return 0;
}
### Prompt

Develop a solution in cpp 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 <iostream>
#include <vector>
#include <string>
#include <map>
#include <algorithm>

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

using namespace std;

int main() {
  while(1) {
    int n,k;
    cin >> n >> k;
    if (!n) break;
    vector<int> v(n+1);
    for (int i = 0; i < k; ++i) {
      int a;
      cin >> a;
      v[a] = 1;
    }
    int l=0;
    int zero = 0;
    int max_len = 0;
    for (int i = 1; i <= n; ++i) {
      if (v[i] == 0) {
        zero++;
      }
      while(zero > v[0]) {
        if(v[++l] == 0) {
          zero--;
        }
      }
      max_len = max(max_len, i - l);
    }
    cout << max_len << endl;
  }
  return 0;
}
```