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>
double v[210];
int main() {
  int n, b;
  while (scanf("%d%d", &n, &b) != EOF) {
    double sum = 0;
    memset(v, 0, sizeof(v));
    for (int i = 1; i <= n; i++) {
      scanf("%lf", &v[i]);
      sum += v[i];
    }
    sum += b;
    int f = 0;
    double ave = sum / n, need = 0;
    for (int i = 1; i <= n; i++) {
      if (v[i] > ave) {
        f = 1;
        break;
      }
      need += ave - v[i];
    }
    if (f || fabs(need - b) > 1e-10) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; i++) printf("%.6lf\n", ave - v[i]);
  }
}