N robbers have robbed the bank. As the result of their crime they chanced to get M golden coins. Before the robbery the band has made an agreement that after the robbery i-th gangster would get Xi/Y of all money gained. However, it turned out that M may be not divisible by Y.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#define N 1005
using namespace std;
typedef long long ll;
int K[N];
double S[N];
struct rec{
    int id;
    double num;
}r[N];
int cmp1(rec a, rec b) {
    return a.num < b.num;
}
int cmp2(rec a, rec b) {
    return a.num > b.num;
};
int main() {
    int n;
    while (scanf("%d", &n) != EOF) {
        int M, Y;
        scanf("%d %d", &M, &Y);
        int X, sum = 0;
        for (int i = 0; i < n; i++) {
            scanf("%d", &X);
            S[i] = (M * X) / Y;
            K[i] = (int)S[i];
            if (1 - (S[i] - K[i]) < S[i] - K[i]) {
                K[i]++;         
            }
            r[i].num = abs((S[i] + 1.0) / M - X * 1.0 / Y) - abs(S[i] * 1.0 / M - X * 1.0 / Y);
            r[i].id = i; 
            sum += K[i];
        }
        if (sum < M) {
            int temp = M - sum;
            sort(r, r + n, cmp1);
            for (int i = 0; i < temp; i++) {
                K[r[i].id]++;
            }
        } else if (sum > M) {
            int temp = sum - M;
            sort(r, r + n, cmp2);
            for (int i = 0; i < temp; i++) {
                K[r[i].id]--;
            }
        }
        printf("%d", K[0]);
        for (int i = 1; i < n; i++) {
            printf(" %d", K[i]);
        }
        printf("\n");
    }
    return 0;
}