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#include <bits/stdc++.h>
using namespace std;

const int MAX_N = 50000;
int N, M;
u_short a[MAX_N];
int counter[129];

u_short d[MAX_N];

void process(){
	int last_element = M + a[0];
	for(int v = 0; v < 129; v++){
		counter[v] = 0;
	}

	int i = 1;
	while(i < N){
		// we assume that counter contains inside all
		// the indices in range [0, i - 2]. We further
		// assume that all a[j], where j in range [0, i - 2]
		// has already been increased in preparation for
		// this new plan. (as well as last_element).

		// sum of all elements.
		int SUM = 0;
		for(int v = 128; v >= 0; v--){
			SUM += v * counter[v];
		}

		// I "must" zero out (i + 1)/2 values.
		int real_amount_zeroed = 0;
		if(last_element > 128){
			real_amount_zeroed++;
		} else {
			SUM += last_element;
			counter[last_element]++;
		}

		// now we find which elements are being zeroed out.
		int last_v = 129;
		int last_zeroed_amt = 0;
		if(real_amount_zeroed >= (i + 1)/2 && SUM <= M){

		} else {
			for(int v = 128; v >= 0; v--){
				// remove the minimal amount needed to satisfy some conditions.
				last_v = v;
				last_zeroed_amt = (i + 1)/2 - real_amount_zeroed;
				if(v > 0){
					last_zeroed_amt = max(last_zeroed_amt, (SUM - M  + v - 1)/v);
				}
				last_zeroed_amt = min(last_zeroed_amt, counter[v]);

				SUM -= v * last_zeroed_amt;
				real_amount_zeroed += last_zeroed_amt;

				// cout << "v = " << v << "  counter = " << counter[v] << "  last_zeroed_amt = " << last_zeroed_amt << endl; 

				if(real_amount_zeroed >= (i + 1)/2 && SUM <= M){
					break;
				}
			}
		}


		int new_i = i + 1;

		// cout << "i = " << i << "  real_amount_zeroed = " << real_amount_zeroed << endl;

		// now we check if we have to "wait" before the plan is accepted.
		if(real_amount_zeroed > (i + 1)/2){
			// in this case we have to wait till this new i.
			while(new_i - real_amount_zeroed < (new_i/2)){
				new_i++;
			}
			new_i++;
			
			// in the case that this new_i is > N, then my
			// resulting plan will throw Nth guy overboard.
			if(new_i > N){
				// revert current additions.
				for(int j = 0; j < i - 1; j++){
					d[j] -= a[j];
				}
				last_element -= a[i - 1];

				// print out the results
				for(int j = N - 1; j >= i; j--){
					cout << -1 << ' ';
				}
				cout << last_element << ' ';
				for(int j = i - 2; j >= 0; j--){
					cout << d[j] << ' ';
				}

				return;
			}
		}

		for(int v = 0; v < 129; v++){
			counter[v] = 0;
		}

		if(last_v <= last_element){
			// in this case I zeroed last_element.
			d[i - 1] = a[i - 1];
			if(last_element == last_v){
				last_zeroed_amt--;
			}
		} else {
			d[i - 1] = last_element + a[i - 1];
		}
		counter[d[i - 1]]++;

		// this loop is the most intense!
		for(int j = i - 2; j >= 0; j--){
			if(d[j] > last_v){
				d[j] = a[j];
			} else if(d[j] == last_v && last_zeroed_amt > 0){
				last_zeroed_amt--;
				d[j] = a[j];
			} else {
				d[j] += a[j];
			}
			counter[d[j]]++;
		}

		for(int j = i; j < new_i - 1; j++){
			d[j] = a[j];
			counter[a[j]]++;
		}

		last_element = M - SUM + a[new_i - 1];

		// for(int j = 0; j < new_i - 1; j++){
		// 	cout << d[j] - a[j] << ' ';
		// }
		// cout << last_element - a[new_i - 1] << endl;

		i = new_i;
	}

	// finally, revert all the changes and print out.
	// revert current additions.
	for(int j = 0; j < N - 1; j++){
		d[j] -= a[j];
	}
	last_element -= a[N - 1];

	cout << last_element << ' ';
	for(int j = N - 2; j >= 0; j--){
		cout << d[j] << ' ';
	}
}

int main(){
	cin.tie(NULL);
	ios_base::sync_with_stdio(0);
	cin >> N >> M;

	for(int i = 0; i < N; i++){
		cin >> a[N - 1 - i];
	}
	
	process();
}