#include <cstdio>
#include <algorithm>
#pragma warning(disable : 4996)
using namespace std;
int n, m, x, c[100009];
int main() {
	while (scanf("%d%d", &n, &m), n) {
		fill(c, c + n, 1);
		int cnt = 0;
		for (int i = 0; i < m; i++) {
			scanf("%d", &x);
			if (x == 0) cnt++;
			else c[--x] = 0;
		}
		int r = 0, sum = 0, ret = 0;
		for (int i = 0; i < n; i++) {
			if (r > i) r = i, sum = 0;
			while (r < n && sum + c[r] <= cnt) sum += c[r++];
			ret = max(ret, r - i);
			sum -= c[i];
		}
		printf("%d\n", ret);
	}
	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 <cstdio>
#include <algorithm>
#pragma warning(disable : 4996)
using namespace std;
int n, m, x, c[100009];
int main() {
	while (scanf("%d%d", &n, &m), n) {
		fill(c, c + n, 1);
		int cnt = 0;
		for (int i = 0; i < m; i++) {
			scanf("%d", &x);
			if (x == 0) cnt++;
			else c[--x] = 0;
		}
		int r = 0, sum = 0, ret = 0;
		for (int i = 0; i < n; i++) {
			if (r > i) r = i, sum = 0;
			while (r < n && sum + c[r] <= cnt) sum += c[r++];
			ret = max(ret, r - i);
			sum -= c[i];
		}
		printf("%d\n", ret);
	}
	return 0;
}
```