#include <iostream>
using namespace std;

int main() {
  while(1) {
    int n, k;
    cin >> n >> k;
    if(n == 0 && k == 0) break;
    int data[n+1];
    fill_n(data, n+1, 0);
    for(int i = 0; i < k; i++) {
      int a;
      cin >> a;
      data[a] = 1;
    }

    int dp[2][n+1];
    fill_n(dp[0], 2*(n+1), 0);
    if(data[0]) dp[1][0] = 1;
    int maxL = -1;
    for(int i = 1; i <= n; i++) {
      if(data[i]) {
	dp[0][i] = dp[0][i-1]+1;
	dp[1][i] = dp[1][i-1]+1;
      } else {
	dp[0][i] = 0;
	dp[1][i] = dp[0][i-1]+1;
      }
      if(maxL == -1 || dp[data[0]][i] > maxL) maxL = dp[data[0]][i];
    }
    cout << maxL << endl;
  }
}
### 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>
using namespace std;

int main() {
  while(1) {
    int n, k;
    cin >> n >> k;
    if(n == 0 && k == 0) break;
    int data[n+1];
    fill_n(data, n+1, 0);
    for(int i = 0; i < k; i++) {
      int a;
      cin >> a;
      data[a] = 1;
    }

    int dp[2][n+1];
    fill_n(dp[0], 2*(n+1), 0);
    if(data[0]) dp[1][0] = 1;
    int maxL = -1;
    for(int i = 1; i <= n; i++) {
      if(data[i]) {
	dp[0][i] = dp[0][i-1]+1;
	dp[1][i] = dp[1][i-1]+1;
      } else {
	dp[0][i] = 0;
	dp[1][i] = dp[0][i-1]+1;
      }
      if(maxL == -1 || dp[data[0]][i] > maxL) maxL = dp[data[0]][i];
    }
    cout << maxL << endl;
  }
}
```