/*input
20
YYGYYYGGGGRGYYGRGRYG
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
typedef long long int ll;
#define ld double
#define pii pair<int,int>
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
const ll maxn=400+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=ll(1e9+7);
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
ll mult(ll a,ll b){
return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
if(b<=0) return 1;
ll res=1LL;
while(b){
if(b&1) res=mult(res,a);
a=mult(a,a);
b>>=1;
}
return res;
}
int arr[maxn];
int dp[maxn][maxn][maxn][3];
int c[3];
int pos[3][maxn];
int cnt[3][maxn];
int main(){
ios::sync_with_stdio(false),cin.tie(0);
int n;
cin>>n;
map<char,int> mp;
mp['R']=0,mp['G']=1,mp['Y']=2;
REP(i,n){
char x;
cin>>x;
int z=mp[x];
REP(j,3) cnt[j][i]=c[j];
pos[z][c[z]]=i;
c[z]++;
}
REP(i,n+1) REP(j,n+1) REP(k,n+1) REP(a,3) dp[i][j][k][a]=INF;
REP(a,3) dp[0][0][0][a]=0;
REP(i,c[0]+1){
REP(j,c[1]+1){
REP(k,c[2]+1){
REP(z,3){
if(z){
int p=pos[0][i];
MNTO(dp[i+1][j][k][0],dp[i][j][k][z]+max(0,cnt[1][p]-j)+max(0,cnt[2][p]-k));
}
if(z!=1){
int p=pos[1][j];
MNTO(dp[i][j+1][k][1],dp[i][j][k][z]+max(0,cnt[0][p]-i)+max(0,cnt[2][p]-k));
}
if(z!=2){
int p=pos[2][k];
MNTO(dp[i][j][k+1][2],dp[i][j][k][z]+max(0,cnt[1][p]-j)+max(0,cnt[0][p]-i));
}
}
}
}
}
int ans=INF;
REP(a,3) MNTO(ans,dp[c[0]][c[1]][c[2]][a]);
if(ans==INF) ans=-1;
cout<<ans;
}
