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
//↓template↓

#include "bits/stdc++.h"
using namespace std;
#define Would
#define you
#define all(n)  n.begin(),n.end()
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
const double pi = acos(-1);
const int SIZE = 1 << 17;
int dx[] = { 1,0,-1,0 }, dy[] = { 0,1,0,-1 }, alp[30];
long long fac[200005], finv[200005], inv[200005];
vector<long long>dij;
struct edge { long long to, cost; };
vector<vector<edge> >G;

long long mod_pow(long long a, long long b) {
	long long res = 1, mul = a;
	for (int i = 0; i < 31; ++i) {
		if (b >> i & 1) {
			res *= mul;
			res %= MOD;
		}
		mul = (mul * mul) % MOD;
	}

	return res;
}

void addedge(int from, int to, int cost) {
	G[from].push_back({ to,cost });
	G[to].push_back({ from,cost });
}

//↑template↑

long long a, b, k[575][575], ans;
int main() {
	cin >> a >> b;
	for (int i = 0; i < a; ++i) {
		for (int j = 0; j < b; ++j) {
			cin >> k[i][j];
			ans += k[i][j] * (i + 1)*(j + 1)*(a - i)*(b - j);
		}
	}
	cout << ans << endl;
}
