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 JAVA):

Solution:
import java.util.*;
import java.math.*;
import java.io.*;
public class Main
	{
	public static void main(String args[])	throws IOException
		{
		Scanner c=new Scanner(System.in);
		int N=c.nextInt();
		int A[]=new int[N];
		int b=c.nextInt();
		for(int i=0;i<N;i++) 
			A[i]=c.nextInt();
		int min=Integer.MAX_VALUE;
		for(int i=0;i<N;i++) 
			{
			if(A[i]<min)
				min=A[i];
			}
		double ans=find_binSearch(A,b,min,200);
		if(ans==-1)
			{
			System.out.println("-1");
			return;
			}
		for(int i=0;i<N;i++) 
			System.out.printf("%.06f\n",ans-A[i]);
		}
	/*
	 * Returns a value v s.t. f(v)>=k. v must lie b/w lo and hi (inclusive).
	 * Take care of overflows which checking f(i)>=k. If f(i) is large for i,
	 * f(i) may overflow. This code avoids all off-by-one errors. Do not give
	 * hi as Long.MAX_VALUE to avoid overflows.
	 */
	private static double find_binSearch(int A[],int b,double lo, double hi)
		{
		//System.out.println("("+lo+","+hi+")");
		if(hi-lo<=(double)(.000000000001))
			{
			double sum=0;
			double mid=(lo+hi)/2;
			for(int i=0;i<A.length;i++) 
				sum+=(mid-A[i]);
			//System.out.println("sum= "+sum);
			if(Math.abs(sum-b)<=.00000001)
				return mid;
			else
				return -1;
			}
		double mid=(lo+hi)/2;
		
		for(int i=0;i<A.length;i++) 
			{
			if(A[i]>mid)
				 return find_binSearch(A,b,mid,hi);
			}
		double sum=0;
		for(int i=0;i<A.length;i++) 
			sum+=(mid-A[i]);
		if(Math.abs(sum-b)<=.00000001)
			return mid;
		if(sum>b)
			return find_binSearch(A,b,lo,mid);
		else
			return find_binSearch(A,b,mid,hi);
		}
	}

//must declare new classes here