Problem: For a given array $a_1, a_2, a_3, ... , a_N$ of $N$ elements and an integer $K$, find the smallest sub-array size (smallest window length) where the elements in the sub-array contains all integers in range [$1, 2, ..., K$]. If there is no such sub-array, report 0.

Constraints

* $1 \leq N \leq 10^5$
* $1 \leq K \leq 10^5$
* $1 \leq a_i \leq 10^5$

Input

The input is given in the following format.

$N$ $K$
$a_1$ $a_2$ ... $a_N$

Output

Print the smallest sub-array size in a line.

Examples

Input

6 2
4 1 2 1 3 5


Output

2


Input

6 3
4 1 2 1 3 5


Output

3


Input

3 4
1 2 3


Output

0

Solution (in CPP):

Solution:
#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
const int INF = 1<<30; 
const int MOD = 1e9 + 7;
const int dy[] = {1, 0, -1, 0};
const int dx[] = {0, 1, 0, -1};
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, k; cin >> n >> k;
    vector<int> v(n);
    for(int i = 0; i < n; i++) cin >> v[i];
    int res = INF;
    int right = 0;
    map<int, int> mp;
    queue<int> que;
    for(int left = 0; left < n; left++)
    {
        while(right < n && mp.size() < k)
        {
            if(v[right] <= k)
            {
                que.push(v[right]);
                mp[v[right]]++;
            }
            right++;
        }
        if(mp.size() == k) res = min(right - left, res);
        if(left == right) right++;
        else if(v[left] == que.front()) 
        {
            int head = que.front(); que.pop();
            mp[head]--;
            if(mp[head] == 0) mp.erase(head);
        }
    }
    cout << (res != INF ? res : 0) << endl;
    return 0;
}