#include<bits/stdc++.h>
#pragma GCC optimize(2)
#define qpow(x) ((x * x * x))
using namespace std;

struct node{
	int val1, val2, val3;
};

int L, Q, x;
node ans[100005];

/*
void solve(int x){
	for (int i = -L; i <= L; i ++){
		for (int j = -L; j <= L; j ++){
			for (int k = -L; k <= L; k ++){
				if (qpow(i) + qpow(j) + qpow(k) == x){
					printf("%d %d %d\n", i, j, k);
					return;
				}
			}
		}
	}
	printf("%d %d %d\n", L + 1, L + 1, L + 1);
}
*/

void __init(){
	for (int i = 1; i <= L; i ++){
		ans[i].val1 = L + 1, ans[i].val2 = L + 1, ans[i].val3 = L + 1;
	}
	for (int i = -L; i <= L; i ++){
		for (int j = -L; j <= L; j ++){
			for (int k = -L; k <= L; k ++){
				int tmp = qpow(i) + qpow(j) + qpow(k);
				if (tmp < 1 || tmp > L) continue;
				ans[tmp].val1 = i, ans[tmp].val2 = j, ans[tmp].val3 = k;
			}
		}
	}
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("../data.in", "r", stdin);
	freopen("../data.out", "w", stdout);
#endif

	scanf("%d %d", &L, &Q);
	__init();
	while (Q --){
		scanf("%d", &x);
		printf("%d %d %d\n", ans[x].val1, ans[x].val2, ans[x].val3);
	}
	
	return 0;
}