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 <bits/stdc++.h>
#define REP(i, a, n) for(ll i=a; i<n; i++)
#define RREP(i, a, n) for(ll i=n-1; i>=a; i--)
typedef long long ll;
const ll mod =1e9+7;
const ll inf =1e18;
using namespace std;

ll h, w;
ll a[501][501];
ll ans;
int main(){
    cin>>h>>w;
    REP(i, 0, h) REP(j, 0, w) cin>>a[i][j];

    REP(i, 0, h){
        REP(j, 0, w){
            if((i==0&&j==0) || (i==0&&j==w-1) || (i==h-1&&j==0) || (i==h-1&&j==w-1)){
                ans+=(a[i][j]*h*w);
            }
            else if(i==0 || i==h-1){
                ans+=(a[i][j]*(j+1)*(w-j)*h);
            }
            else if(j==0 || j==w-1){
                ans+=(a[i][j]*(i+1)*(h-i)*w);
            }
            else{
                ans+=(a[i][j]*(j+1)*(i+1)*(w-j)*(h-i));
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}