#include<cstdio>
using namespace std;
 
#define REP(i,a,b) for(int i=a; i<b; ++i)
#define rep(i,n) REP(i,0,n)
 
int sum[100010];
 
int main() {
  int n,k,t,zero;
  while(scanf("%d %d", &n, &k), n|k) {
    rep(i,n+1) sum[i] = 1; sum[0] = 0;
    zero = 0;
    rep(i,k) {
      scanf("%d", &t);
      if(t == 0) zero++;
      else sum[t] = 0;
    }
 
    rep(i,n) sum[i+1] += sum[i];
 
    int res = 0,start = 0,end = 0;
    while(end <= n) {
      if(sum[end]-sum[start] > zero) start++;
      if(end - start > res) res = end - start;
      end++;
    }
    printf("%d\n",res);
  }
}
### Prompt

Create a solution in cpp for 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<cstdio>
using namespace std;
 
#define REP(i,a,b) for(int i=a; i<b; ++i)
#define rep(i,n) REP(i,0,n)
 
int sum[100010];
 
int main() {
  int n,k,t,zero;
  while(scanf("%d %d", &n, &k), n|k) {
    rep(i,n+1) sum[i] = 1; sum[0] = 0;
    zero = 0;
    rep(i,k) {
      scanf("%d", &t);
      if(t == 0) zero++;
      else sum[t] = 0;
    }
 
    rep(i,n) sum[i+1] += sum[i];
 
    int res = 0,start = 0,end = 0;
    while(end <= n) {
      if(sum[end]-sum[start] > zero) start++;
      if(end - start > res) res = end - start;
      end++;
    }
    printf("%d\n",res);
  }
}
```