Little penguin Polo 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 *a*<sub class="lower-index">*ij*</sub>.

In one move the penguin 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.
The first line contains three integers *n*, *m* and *d* (1 ≤ *n*, *m* ≤ 100, 1 ≤ *d* ≤ 10<sup class="upper-index">4</sup>) — 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 *a*<sub class="lower-index">*ij*</sub> (1 ≤ *a*<sub class="lower-index">*ij*</sub> ≤ 10<sup class="upper-index">4</sup>).
from math import floor, ceil
n, m, d = map(int, input().split())
grid = [None for _ in range(n)]; s = 0
for i in range(n):
	grid[i] = list(map(int, input().split()))
	s += sum(grid[i])
if s % d != 0:
	print(-1)
else:
	count = 0; avg = s / (n * m); flag = True
	if avg % d == 0:
		for i in range(n):
			for j in range(m):
				if grid[i][j] % d != 0:
					flag = False
					break
				count += abs(grid[i][j] - avg) // d
			if not flag:
				break
		if flag:
			print(count)
		else:
			print(-1)
	else:
		a1 = floor(avg / d) * d
		a2 = ceil(avg / d) * d
		count = 0; flag = True
		if a1 % d == 0:
			for i in range(n):
				for j in range(m):
					if grid[i][j] % d != 0:
						flag = False
						break
					count += abs(grid[i][j] - a1) // d
				if not flag:
					break
			if not flag:
				print(-1)
				exit()
		count1 = count
		count = 0; flag = True
		if a2 % d == 0:
			for i in range(n):
				for j in range(m):
					if grid[i][j] % d != 0:
						flag = False
						break
					count += abs(grid[i][j] - a2) // d
				if not flag:
					break
			if not flag:
				print(-1)
				exit()
		print(min(count1, count))
