Polo the Penguin and Matrix

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_ij.

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 a_ij (1 ≤ a_ij ≤ 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

Copy

2 2 2
2 4
6 8

output

Copy

4

input

Copy

1 2 7
6 7

output

Copy

-1
题解：看数据比较小，直接暴力跑了一遍。

#pragma warning(disable:4996)
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 10005;

int n, m, d;
int a[maxn];

int main()
{
    while (cin >> n >> m >> d) {
        int cnt = 0, ma = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                int tp;
                scanf("%d", &tp);
                a[++cnt] = tp;
                ma = max(ma, tp);
            }
        }
        int ans = 2000000007;
        for (int i = 1; i <= ma; i++) {
            int tem = 0;
            bool flag = true;
            for (int j = 1; j <= cnt; j++) {
                if (abs(a[j] - i) % d) { flag = false; break; }
                tem += abs(a[j] - i) / d;
            }
            if (flag) ans = min(ans, tem);
        }
        if (ans == 2000000007) cout << "-1" << endl;
        else cout << ans << endl;
    }
    return 0;
}

正解：
　　a1+x1*d=A;
　　a2+x2*d=A;
　　a3+x3*d=A;
　　·····
显然，a1%d=A%d=a2%d=A%d=a3%d=A%d=···,所以如果矩阵中每个元素模d的余数不想等，则必然不能通过加减使得相等。然后排序，再从中间向两边加减d就行了。

#pragma warning(disable:4996)
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 10005;

int n, m, d;
int a[maxn];

int main()
{
    while (cin >> n >> m >> d) {
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) scanf("%d", &a[++cnt]);
        }

        bool flag = true;
        for (int i = 2; i <= cnt; i++) if (a[i] % d != a[1] % d) { flag = false; break; }

        if (!flag) printf("-1\n");
        else {
            int ans = 0;
            int pos = (cnt % 2 == 0) ? cnt / 2 : cnt / 2 + 1;

            sort(a + 1, a + cnt + 1);
            for (int i = 1; i <= cnt; i++) if (i != pos) ans += (abs(a[i] - a[pos]) / d);
            cout << ans << endl;
        }
    }
    return 0;
}