#include <stdio.h>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <iterator>
#include <assert.h>
#pragma warning(disable:4996) 
 
typedef long long ll;
#define MIN(a, b) ((a)>(b)? (b): (a))
#define MAX(a, b) ((a)<(b)? (b): (a))
#define LINF  9223300000000000000
#define LINF2 1223300000000000000
#define INF 2140000000
const long long MOD = 1000000007;
//const long long MOD = 998244353;

using namespace std;

void solve()
{
    int n, K;
    scanf("%d%d", &n, &K);
    vector<int> p(n);
    vector<pair<int,int> > next(n*30);
    int i;
    for (i = 0; i < n; i++) {
        scanf("%d", &p[i]); p[i]--;
        int ne = (i + p[i] + 1);
        int add = 0;
        if (ne >= n) {
            ne -= n;
            add = 1;
        }
        next[i].first = ne;
        next[i].second = add;
    }

#define ID(i, j) ((i)+(j)*n)
    int k;
    for (k = 1; k < 30; k++) {
        for (i = 0; i < n; i++) {
            int ne  = next[ID(i, k - 1)].first;
            int add = next[ID(i, k - 1)].second;
            int ne2 = next[ID(ne, k - 1)].first;
            int add2 = next[ID(ne, k - 1)].second;
            next[ID(i, k)] = make_pair(ne2, add + add2);
        }
    }

    for (i = 0; i < n; i++) {
        int curr = i; ll add = 0;
        for (k = 0; k < 30; k++) {
            if (K & (1 << k)) {
                int ne1 = next[ID(curr, k)].first;
                int add1 = next[ID(curr, k)].second;
                curr = ne1;
                add += add1;
            }
        }

        ll ans = (ll)add * n + curr;
        printf("%lld\n", ans + 1);
    }



    return;
}

int main(int argc, char* argv[])
{
#if 1
    solve();
#else
    int T; scanf("%d", &T);
    while(T--) {
        solve();
    }
#endif
    return 0;
}