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<cstdio>
#include<cmath>
#include<cstring>
#include<vector>
#include<set>
#include<queue>
#include<cstdlib>
#include<algorithm>

using namespace std;
#define pb push_back
#define mp make_pair
#define ls(x) (x << 1)
#define rs(x) (x<<1|1)
#define SZ(x) (int)(x).size()

typedef long long LL;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<LL, LL> pll;

const double eps = 1e-10;
const double PI = acos(-1.0);
//const double PI = 3.14159265358979323846264338327950288419716939937510
int dcmp(const double &x) { if (fabs(x) < eps) return 0; return x < 0 ? -1 : 1; }
const int MOD = 1e9 + 7;
const int INF = 2e9;
const double INF_d = 1e64;

template<class T>
T qmod(T a, T b){ T ret = 1; while (b){ if (b & 1) ret *= a; b >>= 1; a *= a; } return ret; }
template<class T>
T gcd(T a, T b){ return !b ? a : gcd(b, a % b); }
template<class T>
T ex_gcd(T a, T b, T &x, T &y){
    if (!b){ x = 1, y = 0; return a; }
    T t, ret;
    ret = ex_gcd(b, a % b, x, y);
    t = x, x = y, y = t - a / b * y;
    return ret;
}
template<class T>
T inv(T t, T p){ return t == 1 ? 1 : (p - p / t) * inv(p % t, p) % p; }
// head

const int N = 1e5 + 7;
int n, k, a[N], res, cnt, vis[N];

void solve(){
    res = INF;
    int l, r; 
    l = r = cnt = 0;
    memset(vis, 0, sizeof(vis));
    while (r < n){
        while (cnt < k && r < n){
            if (a[r] >= 1 && a[r] <= k){
                if (!vis[a[r]]) cnt++;
                vis[a[r]]++;
            }
            r++;
        }
        if (cnt < k) break;
        while (cnt >= k && l < r){
            res = min(res, r - l);
            if (a[l] >= 1 && a[l] <= k){
                vis[a[l]]--;
                if (!vis[a[l]]) cnt--;
            }
            l++;
        }
    }
    if (res == INF) res = 0;
    printf("%d\n", res);
}

int main(){
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    solve();
    return 0;
}