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;
template <class T>
inline T MIN(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline T MAX(T a, T b) {
  return a > b ? a : b;
}
template <class T>
inline void getInt(T& x) {
  char c;
  int mul = 1;
  while ((c = getchar()) != EOF) {
    if (c == '-') mul = -1;
    if (c >= '0' && c <= '9') {
      x = c - '0';
      break;
    }
  }
  if (c == EOF) {
    x = EOF;
    return;
  }
  while (c = getchar()) {
    if (c >= '0' && c <= '9') {
      x = (x << 1) + (x << 3) + (c - '0');
    } else
      break;
  }
  x *= mul;
}
int N;
int b;
int amount[101];
int main() {
  cin >> N >> b;
  int maks = -1;
  for (int i = 0, _n = (N); i < _n; i++) {
    cin >> amount[i];
    maks = MAX(maks, amount[i]);
  }
  int need = 0;
  for (int i = 0, _n = (N); i < _n; i++) {
    need += maks - amount[i];
  }
  if (need > b) {
    puts("-1");
    return 0;
  } else {
    double left = b - need;
    for (int i = 0, _n = (N); i < _n; i++) {
      printf("%.6f\n", maks - amount[i] + (left / N));
    }
  }
  return 0;
}