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 <string>
#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <map>
#include <set>
typedef long long ll;
const int mod=1e9+7;
#define lol(i,n) for(int i=0;i<n;i++)
#define tot(i,n) for(int i=0;i<=n;i++)

using namespace std;
int H,W;
int A[501][501];
ll sum=0;
int main() {
    cin >> H >> W;
    for (int i=1;i<=H;i++) {
        for (int j=1;j<=W;j++) {
            cin >> A[i][j];
            ll S1=i*(H+1-i);
            ll S2=j*(W+1-j);
            sum+=A[i][j]*S1*S2;
        }
    }
    cout << sum << endl;
    return 0;
}