// by xcwgf666

#include <cstdio>
#include <cassert>
#include <algorithm>
using namespace std;

int n, k, i, j;
long long a[100000 + 5], mx;

int main (int argc, char * const argv[]) {
	scanf("%d %d", &n, &k);
	assert(1 <= n && n <= 100000 && 0 <= k && k <= 2000000000);
	// one can prove (or just check) that there is a cycle of the size not greater than two
	// so the required number of iterations is actually very small, so we can actually do a brute force
	if (k % 2 == 1) k = 1; else k = min(k, 2); 
	for(i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);
		assert(abs(a[i]) <= 2000000000);
	}
	for(i = 1; i <= k; i++) { // performing what is described in the problem
		mx = a[1];
		for(j = 2; j <= n; j++) mx = max(mx, a[j]);
		for(j = 1; j <= n; j++) a[j] = mx - a[j];
	}
	for(i = 1; i < n; i++) printf("%d ", (int)a[i]);
	printf("%d\n", (int)a[n]);
    return 0;
}