Problem: K: Relief (Angel Relief)

Tenma, an angel, decides to save a city.

The city has a rectangular shape divided into north-south $ H $ parcels x east-west $ W $ parcels, with houses in each parcel.

The $ X $ th section from the north and the $ Y $ th section from the west are represented by $ (X, Y) $.

The house in parcel $ (i, j) $ is inhabited by $ A_ {i, j} $ people.

Tenshin chooses a rectangular area whose sides are parallel or vertical to north-south or east-west, and rescues all the people living in it one by one.

Tenshin does this for every possible rectangle.

Find the total number of times Tenma-san rescues people.

input

The integers $ H, W $ are given on the first line, separated by blanks.

Of the following $ H $ lines, the integers $ A_ {i, 1}, A_ {i, 2}, A_ {i, 3}, \ dots, A_ {i, W} $ are blank on the $ i $ line. Given as a delimiter.

output

Output the total number of times Tenma-san rescues people.

Constraint

* $ H, W $ are integers between $ 1 $ and $ 500 $
* $ A_ {i, j} $ are all integers greater than or equal to $ 1 $ and less than or equal to $ 9 $.



Input example 1


twenty two
1 2
4 8


Output example 1


60


For example, if you choose a rectangular area with $ (1, 1) $ in the upper left and $ (2, 2) $ in the lower right, you will rescue $ 15 $ people there one by one, so a total of $ 15 $ times. Relief.

$ 1, 2, 3, 4, 5, 8, 10, 12, 15 $ relief for each of the $ 9 $ rectangular areas, for a total of $ 60 $.

Input example 2


twenty three
one two Three
4 5 6


Output example 2


140






Example

Input

2 2
1 2
4 8


Output

60

Solution (in CPP):

Solution:
#include <iostream>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <stdio.h>
#include <vector>
#define int long long
#define endre getchar();getchar();return 0
using namespace std;
bool prime(int n) {
	for (int i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}
int gcd(int x, int y) {
	if (x < y)swap(x, y);
	if (y == 0)return x;
	return (y, x%y);
}
int lcm(int x, int y) {
	return x * y / gcd(x, y);
}
int kai(int n) {
	if (n == 0)return 1;
	return kai(n - 1)*n;
}
int mod_pow(int x, int y, int mod) {
	int res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x%mod;
		}
		x = x * x%mod;
		y >>= 1;
	}
	return res;
}
int n, m, a[505][505], cnt;
signed main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			cin >> a[i][j];
			cnt += a[i][j] * (i*(n - i + 1))*(j*(m - j + 1));
		}
	}
	cout << cnt << endl;
	endre;
}