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

int main() {
    int n, k;
    while(cin >> n >> k, n) {
        vector<int> a(k);
        vector<int> s(n + 2, 1);
        s[0] = s.back() = 0;
        int w = 0;
        for(int i = 0; i < k; ++i) {
            cin >> a[i];
            if(a[i] == 0) {
                w = 1;
            } else {
                s[a[i]] -= 1;
            }
        }
        for(int i = 0; i < n + 1; ++i) {
            s[i + 1] += s[i];
        }

        int ans = 0;
        for(int i = 0; i < n; ++i) {
            auto check = [&] (int x) {
                return s[x] - s[i] <= w;
            };
            int lb = i, ub = n + 1;
            while(ub - lb > 1) {
                const int mid = (ub + lb) >> 1;
                (check(mid) ? lb : ub) = mid;
            }
            ans = max(ans, ub - i - 1);
        }
        cout << ans << endl;
    }
}

