#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int H, W;
cin >> H >> W;
vector<string> grid(H);
for (int i = 0; i < H; i++){
cin >> grid[i];
}
// Precompute for each row: rightO[i][j] = number of 'O's in row i in columns j+1...W-1.
vector<vector<int>> rightO(H, vector<int>(W, 0));
for (int i = 0; i < H; i++){
// Last column: no cell to its right.
rightO[i][W-1] = 0;
for (int j = W - 2; j >= 0; j--){
rightO[i][j] = rightO[i][j+1] + (grid[i][j+1] == 'O');
}
}
// Precompute for each column: downI[i][j] = number of 'I's in column j in rows i+1...H-1.
// We use a 2D array downI with same dimensions.
vector<vector<int>> downI(H, vector<int>(W, 0));
for (int j = 0; j < W; j++){
// Last row: no cell below.
downI[H-1][j] = 0;
for (int i = H - 2; i >= 0; i--){
downI[i][j] = downI[i+1][j] + (grid[i+1][j] == 'I');
}
}
// Count the number of quadruplets.
long long ans = 0;
for (int i = 0; i < H; i++){
for (int j = 0; j < W; j++){
if (grid[i][j] == 'J') {
// For top-left (i,j) with a jewel, multiply
// the number of orbs to its right in the same row and
// the number of ingots below in the same column.
ans += (long long) rightO[i][j] * downI[i][j];
}
}
}
cout << ans << "\n";
return 0;
}
