Star and Matrix



Star has an n×m matrix, consisting of integers. Let's index the matrix rows from 1 to n from top to bottom and let's index the columns from 1 to m from left to right. Let's represent the matrix element on the intersection of row i and column j as aij.

 

In one move the Star can add or subtract number d from some matrix element. Find the minimum number of moves needed to make all matrix elements equal. If the described plan is impossible to carry out, say so.

 

Input

 The first line contains three integers n, m and d (1≤n,m≤100,1≤d≤10⁴) — the matrix sizes and the d parameter. Next n lines contain the matrix: the j-th integer in the i-th row is the matrix element aij (1≤aij≤10⁴).

Output

 In a single line print a single integer — the minimum number of moves the penguin needs to make all matrix elements equal. If that is impossible, print "-1" (without the quotes).

Sample Input

2 2 2

2 4

6 8

 

1 2 7

6 7

Sample Output

4

-1

做法：找出这个矩阵的中位数，然后从两边开始，分别减去中位，再除以d，如果不能被整除就跳出循环，输出-1，不然就继续计算

#include<iostream>
#include<algorithm>

using namespace std;

int arr[1000000];

int main()
{
	int n,m,k;
	cin >> n >> m >> k;
	for(int i=0; i < n*m; i++)
		cin >> arr[i];
	sort(arr,arr+n*m);
	int pivot=(n*m)/2;
	int count=0,o=0;
	for(int i=0; i < pivot; i++){
		if((arr[pivot]-arr[i])%k==0){
			count+=(arr[pivot]-arr[i])/k;
		}
		else{
			o=1;
			break;
		}
	}
	if(o!=1){
	for(int i=n*m-1; i >pivot; i--){
		if((arr[i]-arr[pivot])%k==0){
			count+=(arr[i]-arr[pivot])/k;
		}
		else{
			o=1;
			break;
		}
	}
	
	};
	if(o==1) cout << "-1" << endl;
	else cout << count << endl;
}