#include <iostream>
#include <vector>
#include <string>
using namespace std;
/**
* JOI Bitaro the Brave yechimi.
* Vaqt murakkabligi: O(H * W)
* Xotira murakkabligi: O(H * W)
*/
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int H, W;
cin >> H >> W;
vector<string> S(H);
for (int i = 0; i < H; i++) {
cin >> S[i];
}
// Har bir (i, j) dan o'ngda nechta 'O' borligini saqlash
vector<vector<int>> countO(H, vector<int>(W, 0));
// Har bir (i, j) dan pastda nechta 'I' borligini saqlash
vector<vector<int>> countI(H, vector<int>(W, 0));
// O larni qatorlar bo'yicha o'ngdan chapga sanaymiz
for (int i = 0; i < H; i++) {
int currentO = 0;
for (int j = W - 1; j >= 0; j--) {
if (S[i][j] == 'O') currentO++;
countO[i][j] = currentO;
}
}
// I larni ustunlar bo'yicha pastdan yuqoriga sanaymiz
for (int j = 0; j < W; j++) {
int currentI = 0;
for (int i = H - 1; i >= 0; i--) {
if (S[i][j] == 'I') currentI++;
countI[i][j] = currentI;
}
}
long long totalPower = 0;
for (int i = 0; i < H; i++) {
for (int j = 0; j < W; j++) {
if (S[i][j] == 'J') {
// J joylashgan (i, j) uchun o'ngdagi O lar va pastdagi I larni ko'paytiramiz
totalPower += (long long)countO[i][j] * countI[i][j];
}
}
}
cout << totalPower << endl;
return 0;
}
