#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define ALL(A) A.begin(), A.end()
#define FOR(i, a, b) for(int i = a; i <= (int)b; i++)
#define FORD(i, a, b) for(int i = a; i >= (int)b; i--)
#define taskname "kieuoanh"
const int N = 5e5 + 5;
int n;
string s;
int dp[405][405][405][3];
vector <int> preR, preG, preY;
int calc(int l, int r, vector <int> v){
int cnt = 0;
for(int i = l; i < (int)v.size() && v[i] <= r; i++) cnt++;
return cnt;
}
void Solve(){
cin >> n >> s; s = "*" + s;
FOR(i, 1, n){
if(s[i] == 'R') preR.push_back(i);
if(s[i] == 'G') preG.push_back(i);
if(s[i] == 'Y') preY.push_back(i);
}
memset(dp, 0x3f, sizeof dp);
int n = preR.size(), m = preG.size(), p = preY.size();
dp[0][0][0][0] = dp[0][0][0][1] = dp[0][0][0][2] = 0;
FOR(i, 0, n){
FOR(j, 0, m) FOR(k, 0, p){
FOR(l, 0, 2){
int &cur = dp[i][j][k][l];
if(l != 0 && i + 1 <= n){
dp[i + 1][j][k][0] = min(dp[i + 1][j][k][0], cur + calc(j, preR[i], preG) + calc(k, preR[i], preY));
}
if(l != 1 && j + 1 <= m){
dp[i][j + 1][k][1] = min(dp[i][j + 1][k][1], cur + calc(i, preG[j], preR) + calc(k, preG[j], preY));
}
if(l != 2 && k + 1 <= p){
dp[i][j][k + 1][2] = min(dp[i][j][k + 1][2], cur + calc(i, preY[k], preR) + calc(j, preY[k], preG));
}
}
}
}
int ans = 2e9;
FOR(l, 0, 2){
ans = min(ans, dp[n][m][p][l]);
}
cout << (ans >= 1e9 ? - 1 : ans);
}
signed main(){
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
if(fopen(taskname".inp", "r")){
freopen(taskname".inp", "r", stdin);
freopen(taskname".out", "w", stdout);
}
Solve();
return 0;
}
