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 aij.

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.

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).

Examples

Input

2 2 2
2 4
6 8


Output

4


Input

1 2 7
6 7


Output

-1
# https://codeforces.com/problemset/problem/289/B

"""
There is an nxm matrix with integer entries.
In a move the penguin can add or subtract a number d from some matrix element
Find the min number of moves needed to make all entries equal.

First the structure of a matrix is in not important
What is important is that all entries are the same modulo d

We can add or subtract d. We can argue that the value that
will yield the least number of moves must exist within the range
of the current values
"""


def main():
    n, m, d = map(int, input().split())

    entries = []
    for i in range(n):
        entries.extend(list(map(int, input().split())))

    residues = [entry % d for entry in entries]
    if residues.count(residues[0]) != len(residues):
        return -1

    entries.sort()
    median = entries[(m*n)//2]

    min_moves = sum([abs(entry - median) // d for entry in entries])  # Clearly an upper bound

    return min_moves


if __name__ == '__main__':
    print(main())

