/*
    IN THE NAME OF GOD
*/
#include <bits/stdc++.h>

// #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")

using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef long double ld;

#define F                                      first
#define S                                      second
#define Mp                                     make_pair
#define pb                                     push_back
#define pf                                     push_front
#define size(x)                                ((ll)x.size())
#define all(x)                                 (x).begin(),(x).end()
#define kill(x)		                           cout << x << '\n', exit(0);
#define fuck(x)                                cout << "(" << #x << " , " << x << ")" << endl
#define endl                                   '\n'

const int N = 3e5+23, lg = 18;
ll Mod = 1e9+7; //998244353;

inline ll MOD(ll a, ll mod=Mod) {a%=mod; (a<0)&&(a+=mod); return a;}
inline ll poww(ll a, ll b, ll mod=Mod) {
    ll ans = 1;
    a=MOD(a, mod);
    while (b) {
        if (b & 1) ans = MOD(ans*a, mod);
        b >>= 1;
        a = MOD(a*a, mod);
    }
    return ans;
}

ll ans, n, deg[2][N], y[10];
bitset<2002> b[2][2002];
int main () {
	ios_base::sync_with_stdio(false), cin.tie(0);

	cin>>n;
	for(int i=1; i<=n; i++) {
		for(int j=1; j<=n; j++) {
			char ch; cin>>ch;
			int x = (ch=='Y');
			if(j!=i) {
				b[x][i][j] = 1;
				deg[x][i] ++;
			}
		}
	}

	ans = -n*(n-1)*(n-2)*(n-3)/8ll;

	ll ans3 = 0;
	for(int i=1; i<=n; i++) {
		ans3 += (2ll*deg[0][i]*deg[1][i]*(n-3));
	}
	ans3/=(4ll); ans+=ans3;
	//fuck(ans);
	ll ans2 = 0;
	for(int i=1; i<=n; i++) {
		for(int j=1; j<=i; j++) {
			if(i==j) continue;
			if(b[1][i][j] == 1) {
				ll num = (b[1][i]&b[1][j]).count();
				ans2 += (2ll*num*(num-1));
				num = (b[0][i]&b[0][j]).count();
				ans2 -= (2ll*num*(num-1));
			} else {
				ll num = (b[0][i]&b[0][j]).count();
				ans2 += (2ll*num*(num-1));
				num = (b[1][i]&b[1][j]).count();
				ans2 -= (2ll*num*(num-1));
			}
		}
	}
	//ans = ans - (y[1]+y[2]+y[3]+y[4])*(n-3);
	ans2/=(8ll);

	cout<<ans+ans2<<endl;

	return 0;
}