Problem: For a given array $a_1, a_2, a_3, ... , a_N$ of $N$ elements and an integer $L$, find the minimum of each possible sub-arrays with size $L$ and print them from the beginning. For example, for an array $\\{1, 7, 7, 4, 8, 1, 6\\}$ and $L = 3$, the possible sub-arrays with size $L = 3$ includes $\\{1, 7, 7\\}$, $\\{7, 7, 4\\}$, $\\{7, 4, 8\\}$, $\\{4, 8, 1\\}$, $\\{8, 1, 6\\}$ and the minimum of each sub-array is 1, 4, 4, 1, 1 respectively.

Constraints

* $1 \leq N \leq 10^6$
* $1 \leq L \leq 10^6$
* $1 \leq a_i \leq 10^9$
* $L \leq N$

Input

The input is given in the following format.

$N$ $L$
$a_1$ $a_2$ ... $a_N$

Output

Print a sequence of the minimum in a line. Print a space character between adjacent elements.

Example

Input

7 3
1 7 7 4 8 1 6


Output

1 4 4 1 1

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 = 1e6 + 7;
int n, L, a[N];
vector<int> res;

void solve(){
    deque<int> dq; while (!dq.empty()) dq.pop_back();
    for (int i = 0; i < n; i++){
        while (SZ(dq) && a[dq.back()] >= a[i]){
            dq.pop_back();
        }
        dq.push_back(i);
        int cpr = dq.back() - dq.front();
        while (cpr >= L){
            dq.pop_front(); cpr--;
        }
        if (i >= L - 1){
            res.pb(a[dq.front()]);
        }
    }
    for (int i = 0; i < SZ(res); i++){
        if (i) printf(" ");
        printf("%d", res[i]);
    }
    puts("");
}

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