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 <bits/stdc++.h>

using namespace std;

#define EACH(i,a) for (auto& i : a)
#define FOR(i,a,b) for(int i=(int)a;i<(int)b;++i)
#define RFOR(i,a,b) for(int i=(int)b-1;i>=(int)a;--i)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define ALL(a) (a).begin(),(a).end()
#define debug(x) cerr << #x << ":" << x << endl;
#define OK(ok) cout << (ok ? "Yes" : "No") << endl;
typedef long long ll;

void CINT(){}
template <class Head,class... Tail>
void CINT(Head&& head,Tail&&... tail) {
  cin >> head; CINT(move(tail)...);
}
#define CIN(...) int __VA_ARGS__;CINT(__VA_ARGS__)
#define LCIN(...) ll __VA_ARGS__;CINT(__VA_ARGS__)
#define SCIN(...) string __VA_ARGS__;CINT(__VA_ARGS__)

const int INF = 1e9 + 1;
const int MOD = 1e9 + 7;
const int MAX_N = 1e5 + 1;

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  CIN(N, K);
  vector< int > a(N);
  REP(i, N) {
    cin >> a[i];
  }
  map< int, int > mp;
  int ans = INF;
  int ub = 0;
  int sm = 0ll;
  REP(lb, N) {
    while (ub < N) {
      if (sm >= K) {
	break;
      }
      if (a[ub++] <= K && mp[a[ub - 1]]++ == 0) {
	sm++;
      }
    }
    if (sm >= K) {
      ans = min(ans, ub - lb);
    }
    /*
    debug(ans);
    debug(lb);
    debug(ub);
    debug(sm);
    cout << endl;
    */
    if (a[lb] <= K && --mp[a[lb]] == 0) {
      sm--;
    }
  }

  if (ans == INF) ans = 0;
  cout << ans << endl;

  

  return 0;
}