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;
const int N = 100;
const long long mod = 1e9 + 7;
vector<int> e[N];
int vis[N];
void dfs(int i, int p, int count) {}
bool comp(pair<long long, long long>& a, pair<long long, long long>& b) {
  return a.first > b.first && a.second < b.second;
}
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  if (a == 0) return 0;
  if (b == 1) return a % mod;
  long long g = (binpow(a, b / 2) % mod) * (binpow(a, b / 2) % mod);
  g %= mod;
  if (b & 1)
    return a * g;
  else
    return g;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    int n, b;
    cin >> n >> b;
    int a[n], g = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      b += a[i];
      g = max(g, a[i]);
    }
    long double c = (long double)b / (long double)n;
    if (g > c)
      cout << -1 << endl;
    else {
      for (int i = 0; i < n; i++) cout << setprecision(20) << c - a[i] << endl;
    }
  }
}