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
#include <bits/stdc++.h>
using namespace std;
bool isprime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<long long int> prime;
void sieve(int n) {
  bool bakh[n + 1];
  memset(bakh, true, sizeof(bakh));
  for (int p = 2; p * p <= n; p++) {
    if (bakh[p] == true) {
      for (int i = p * p; i <= n; i += p) bakh[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (bakh[p]) prime.push_back(p);
}
long long int eulertotient(long long int z) {
  long long int fac = z;
  for (long long int i = 0; prime[i] * prime[i] <= z; i++) {
    if (z % prime[i] == 0) {
      fac -= (fac / prime[i]);
      while (z % prime[i] == 0) z /= prime[i];
    }
  }
  if (z > 1) fac -= (fac / z);
  return fac;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  long long int g = gcd(a, b);
  long long int ans = (a * b) / g;
  return ans;
}
long long int modInverse(long long int a, long long int m) {
  long long int hvf = gcd(a, m);
  if (hvf == 1) return power(a, m - 2, m);
  return -1;
}
void multiply(long long int F[2][2], long long int M[2][2]) {
  long long int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
  long long int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
  long long int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
  long long int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
  F[0][0] = x;
  F[0][1] = y;
  F[1][0] = z;
  F[1][1] = w;
}
void powermat(long long int F[2][2], long long int n) {
  if (n == 0 || n == 1) return;
  long long int M[2][2] = {{1, 1}, {1, 0}};
  powermat(F, n / 2);
  multiply(F, F);
  if (n % 2 != 0) multiply(F, M);
}
long long int fib(long long int n) {
  long long int F[2][2] = {{1, 1}, {1, 0}};
  if (n == 0) return 0;
  powermat(F, n - 1);
  return F[0][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, m, d;
  cin >> n >> m >> d;
  long long int ar[n * m];
  cin >> ar[0];
  long long int r = ar[0] % d;
  bool flag = true;
  for (long long int i = 1; i < n * m; i++) {
    cin >> ar[i];
    if (ar[i] % d != r) flag = false;
  }
  if (!flag) {
    cout << -1 << endl;
    return 0;
  }
  sort(ar, ar + n * m);
  long long int ans = 0, y, k = ar[(n * m) / 2];
  for (long long int i = 0; i < n * m; i++) {
    y = abs(ar[i] - k);
    ans += y / d;
  }
  cout << ans << endl;
}
