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;
using LL = long long;
using ULL = unsigned long long;
const double PI = acos(-1);
template<class T>constexpr T INF() { return ::std::numeric_limits<T>::max(); }
template<class T>constexpr T HINF() { return INF<T>() / 2; }
template <typename T_char>T_char TL(T_char cX) { return tolower(cX); };
template <typename T_char>T_char TU(T_char cX) { return toupper(cX); };
typedef pair<LL, LL> pii;
const int vy[] = { -1, -1, -1, 0, 1, 1, 1, 0 }, vx[] = { -1, 0, 1, 1, 1, 0, -1, -1 };
const int dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };
int popcnt(unsigned long long n) { int cnt = 0; for (int i = 0; i < 64; i++)if ((n >> i) & 1)cnt++; return cnt; }
int d_sum(LL n) { int ret = 0; while (n > 0) { ret += n % 10; n /= 10; }return ret; }
int d_cnt(LL n) { int ret = 0; while (n > 0) { ret++; n /= 10; }return ret; }
LL gcd(LL a, LL b) { if (b == 0)return a; return gcd(b, a%b); };
LL lcm(LL a, LL b) { LL g = gcd(a, b); return a / g*b; };
# define ALL(qpqpq)           (qpqpq).begin(),(qpqpq).end()
# define UNIQUE(wpwpw)        sort(ALL((wpwpw)));(wpwpw).erase(unique(ALL((wpwpw))),(wpwpw).end())
# define LOWER(epepe)         transform(ALL((epepe)),(epepe).begin(),TL<char>)
# define UPPER(rprpr)         transform(ALL((rprpr)),(rprpr).begin(),TU<char>)
# define FOR(i,tptpt,ypypy)   for(LL i=(tptpt);i<(ypypy);i++)
# define REP(i,upupu)         FOR(i,0,upupu)
# define INIT                 std::ios::sync_with_stdio(false);std::cin.tie(0)
# pragma warning(disable:4996)

int main() {
	int N, K, ans = 0, k = 0;
	cin >> N >> K;
	vector<int>v(N), b(K + 1);
	for (int i = 0; i < N; i++)cin >> v[i];
	int s = 0, t = 0;
	for (;;) {
		for (; t<N&&k<K; t++) {
			if (v[t] <= K&&++b[v[t]] == 1)k++;
		}
		if (k<K)break;
		for (; s<N&&k >= K; s++) {
			if (!ans || ans>t - s)ans = t - s;
			if (v[s] <= K && !--b[v[s]])k--;
		}
	}
	cout << ans << endl;
}