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
"""
    Template written to be used by Python Programmers.
    Use at your own risk!!!!
    Owned by adi0311(rating - 5 star at CodeChef and Specialist at Codeforces).
"""
import sys
import bisect
import heapq
from math import *
from collections import defaultdict as dd  # defaultdict(<datatype>) Free of KeyError.
from collections import deque  # deque(list) append(), appendleft(), pop(), popleft() - O(1)
from collections import Counter as c  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb
from bisect import bisect_left as bl, bisect_right as br, bisect
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(var)
def l(): return list(map(int, data().split()))
def sl(): return list(map(str, data().split()))
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [[val for i in range(n)] for j in range(m)]


n, m, d = sp()
mat = list()
for i in range(n):
    arr = l()
    for j in range(m):
        mat.append(arr[j])
if len(set(mat)) == 1:
    out("0")
    exit()
mat.sort()
count = set(mat[:])
for k in count:
    answer = 0
    ele = k
    for i in mat:
        if i == ele:
            continue
        if i > ele:
            if (i-ele) % d != 0:
                out("-1")
                exit()
            answer += (i-ele) // d
        else:
            if (ele-i) % d != 0:
                out("-1")
                exit()
            answer += (ele - i) // d
    mod = min(mod, answer)
out(str(mod))
exit()
