Problem: A group of n merry programmers celebrate Robert Floyd's birthday. Polucarpus has got an honourable task of pouring Ber-Cola to everybody. Pouring the same amount of Ber-Cola to everybody is really important. In other words, the drink's volume in each of the n mugs must be the same.

Polycarpus has already began the process and he partially emptied the Ber-Cola bottle. Now the first mug has a1 milliliters of the drink, the second one has a2 milliliters and so on. The bottle has b milliliters left and Polycarpus plans to pour them into the mugs so that the main equation was fulfilled.

Write a program that would determine what volume of the drink Polycarpus needs to add into each mug to ensure that the following two conditions were fulfilled simultaneously: 

  * there were b milliliters poured in total. That is, the bottle need to be emptied; 
  * after the process is over, the volumes of the drink in the mugs should be equal. 

Input

The first line contains a pair of integers n, b (2 ≤ n ≤ 100, 1 ≤ b ≤ 100), where n is the total number of friends in the group and b is the current volume of drink in the bottle. The second line contains a sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the current volume of drink in the i-th mug.

Output

Print a single number "-1" (without the quotes), if there is no solution. Otherwise, print n float numbers c1, c2, ..., cn, where ci is the volume of the drink to add in the i-th mug. Print the numbers with no less than 6 digits after the decimal point, print each ci on a single line. Polycarpus proved that if a solution exists then it is unique.

Russian locale is installed by default on the testing computer. Make sure that your solution use the point to separate the integer part of a real number from the decimal, not a comma.

Examples

Input

5 50
1 2 3 4 5


Output

12.000000
11.000000
10.000000
9.000000
8.000000


Input

2 2
1 100


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int MAX(int a, int b) { return (a > b) ? a : b; }
double FMAX(double a, double b) { return (a > b) ? a : b; }
int MIN(int a, int b) { return (a < b) ? a : b; }
double FMIN(double a, double b) { return (a < b) ? a : b; }
int GCD(int a, int b) { return b ? GCD(b, a % b) : a; }
int UGCD(unsigned long long int a, unsigned long long int b) {
  return b ? GCD(b, a % b) : a;
}
int LCM(int a, int b) { return a / GCD(a, b) * b; }
void fswop(double &a, double &b) {
  double c = a;
  a = b;
  b = c;
}
void swop(int &a, int &b) {
  a = a ^ b;
  b = a ^ b;
  a = a ^ b;
}
long long int u(long long int a) {
  if (a >= 0) return a;
  return a * (-1);
}
int main() {
  int n, i, k, mas[150], mar[150] = {0}, max = 0;
  scanf("%d %d", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%d", &mas[i]);
    if (mas[i] > max) max = mas[i];
  }
  for (i = 0; i < n; i++) {
    mar[i] = max - mas[i];
    k -= (max - mas[i]);
    mas[i] = max;
    if (k < 0) {
      printf("-1");
      return 0;
    }
  }
  for (i = 0; i < n; i++) {
    printf("%lf\n", (double)mar[i] + (double)k / (double)n);
  }
  return 0;
}