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
#include <bits/stdc++.h>
using namespace std;

#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define abs(a) max((a),-(a))
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define repe(i,n) rep(i,(n)+1)
#define per(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define pere(i,n) rep(i,(n)+1)
#define all(x) (x).begin(),(x).end()
#define SP <<" "<<
#define RET return 0
#define MOD 1000000007
#define INF 1000000000000000000

typedef long long LL;
typedef long double LD;

int main(){
  LL h,w;
  cin >> h >> w;
  vector<vector<int>> a(h,vector<int>(w));
  LL ans=0;
  for(LL i=0;i<h;i++){
    for(LL j=0;j<w;j++){
      cin >> a[i][j];
      ans+=a[i][j]*(i+1)*(h-i)*(j+1)*(w-j);
    }
  }
  cout << ans << endl;
  return 0;
}

