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 = 1e6 + 123;
const double PI = acos(-1);
const long long MOD = 1000000007;
long long dx[] = {+1, 0, -1, 0, +1, +1, -1, -1};
long long dy[] = {0, +1, 0, -1, +1, -1, +1, -1};
long long dx8[] = {+1, +1, -1, -1, +2, +2, -2, -2};
long long dy8[] = {+2, -2, +2, -2, +1, -1, +1, -1};
long long powmod(long long a, long long b) {
  a %= MOD;
  if (!a) return 0;
  long long pr = 1;
  while (b > 0) {
    if (b & 1) {
      pr *= a;
      pr %= MOD;
      --b;
    }
    a *= a;
    a %= MOD;
    b /= 2;
  }
  return pr;
}
long long modinverse(long long a) { return powmod(a, MOD - 2); }
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 or n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i += 6) {
    if (n % i == 0 or n % (i + 2) == 0) return false;
  }
  return true;
}
void seive(bool a[]) {
  long long mx = sqrt(N), ii, jj;
  for (ii = 3; ii <= mx; ii += 2)
    if (!a[ii])
      for (jj = ii * ii; jj < N; jj += 2 * ii) a[jj] = true;
}
void numofdiv(long long a[]) {
  long long mx = sqrt(N), ii, jj;
  for (ii = 1; ii <= mx; ii++) {
    for (jj = ii * ii; jj < N; jj += ii) {
      if (jj == ii * ii)
        a[jj]++;
      else
        a[jj] += 2;
    }
  }
}
void sumofdiv(long long a[]) {
  long long mx = sqrt(N), ii, jj;
  for (ii = 1; ii <= mx; ii++) {
    for (jj = ii * ii; jj < N; jj += ii) {
      if (jj == ii * ii)
        a[jj] += ii;
      else
        a[jj] += ii + jj / ii;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, b, i, req = 0, mx = 0;
  cin >> n >> b;
  vector<int> mug(n);
  for (i = 0; i < n; i++) {
    cin >> mug[i];
    if (mug[i] > mx) mx = mug[i];
  }
  for (i = 0; i < n; i++) req = req + mx - mug[i];
  if (req > b)
    cout << -1 << endl;
  else {
    vector<float> ans(n);
    for (i = 0; i < n; i++) ans[i] = (float(b - req) / float(n)) + mx - mug[i];
    for (auto i : ans) cout << fixed << setprecision(6) << i << endl;
  }
  return 0;
}