// Be nam KHODA
#include<bits/stdc++.h>
using namespace std;
#define all(x) (x).begin(), (x).end()
typedef pair<int, int> pii;
const int N = 400 + 10, inf = 1e9 + 10;
int n;
int a[N];
int dp[N][N][N][3];
int cnt[N][3];
int pd[N][N][3];
int main(){
cin.tie(0), ios::sync_with_stdio(0);
cin >> n;
for(int i = 0; i < n; i++){
char c;
cin >> c;
if(c == 'R')
a[i] = 0;
else if(c == 'G')
a[i] = 1;
else a[i] = 2;
}
for(int j = 0; j < 3; j++){
int cn = 0;
for(int i = 0; i < n; i++){
if(a[i] == j){
cn++;
cnt[cn][j] = i;
}
}
}
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
for(int z = i; z <= j; z++)
for(int d = 0; d < 3; d++)
if(a[z] == d)
pd[i][j][d]++;
int z = count(a, a + n, 0);
int t = count(a, a + n, 1);
int tw = count(a, a + n, 2);
memset(dp, 63, sizeof dp);
dp[0][0][0][0] = dp[0][0][0][1] = dp[0][0][0][2] = 0;
for(int i = 0; i <= z; i++)
for(int j = 0; j <= t; j++)
for(int z = 0; z <= tw; z++)
for(int q = 0; q < 3; q++){
int ar[3] = {i, j, z};
if(ar[q] == 0 or i + j + z == 0){
continue;
}
int mx = max({cnt[i][0], cnt[j][1], cnt[z][2]});
int qq = a[mx];
int op = 1 ^ 2 ^ q ^ qq;
int cost = mx - cnt[ar[q]][q], want = max(0, ar[op] - pd[0][cnt[ar[q]][q]][op]);
cost -= pd[cnt[ar[q]][q]][mx][op] - want + pd[cnt[ar[q]][q] + 1][mx][q];
ar[q]--;
if(q == qq) cost = 0;
for(int qq = 0; qq < 3; qq++) if(qq != q){
dp[i][j][z][q] = min(dp[i][j][z][q], dp[ar[0]][ar[1]][ar[2]][qq] + cost);
}
}
int ans = inf;
for(int i = 0; i < 3; i++)
ans = min(ans, dp[z][t][tw][i]);
if(ans >= inf) ans = -1;
cout << ans << endl;
return 0;
}
