//严格来说，题解中的矩形最值查询其实是二维前缀最值查询，因此可以写一手树状数组套线段树
/*
41 2 2
BRRRBRRRBBBBRRRBBRRRRRRRBBRRRBBBBRRRBRRRB

8

BRRR
BRRR
BBBB
RRRB
BRRR
RRRR
BBRR
RBBB
BRRR
BRRR
B
*/
#include<bits/stdc++.h>
#include<ext/pb_ds/hash_policy.hpp>
#include<ext/pb_ds/assoc_container.hpp>
#ifdef ONLINE_JUDGE
	#define getchar()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
	char buf[1<<21],*p1=buf,*p2=buf;
#endif
using namespace std;using namespace __gnu_pbds;int read(){int num=0,f=1;char c;while(!isdigit(c=getchar()))if(c=='-')f=-1;while(isdigit(c))num=num*10+(c&15),c=getchar();return num*f;}void write(int x,char ch=' '){int F[20],cnt=0;if(!x)putchar('0');if(x<0)putchar('-'),x=-x;while(x)F[cnt++]=x%10+'0',x/=10;while(cnt)putchar(F[--cnt]);putchar(ch);}
namespace Main
{
    const int N=200010,M=N<<1,V=350*N;int n,m,q,k,lim,a[N],b[N],f[N],ans,x,y,z,xx,yy,zz,sa[N],sb[N],s[V][2],v[V],tot;gp_hash_table<int,int>rt[N];char ch[N];void New(int&x){if(!x)v[x=++tot]=-1e9;}void pushup(int k){v[k]=max(v[s[k][0]],v[s[k][1]]);}
    void upd(int&k,int l,int r,int p,int z){New(k),v[k]=max(v[k],z);if(l<r){int mid=(l+r)>>1;if(p<=mid)upd(s[k][0],l,mid,p,z);else upd(s[k][1],mid+1,r,p,z);pushup(k);}}
    int qry(int k,int l,int r,int p){if(!k||p>=r)return v[k];int mid=(l+r)>>1;return (p<=mid?qry(s[k][0],l,mid,p):max(v[s[k][0]],qry(s[k][1],mid+1,r,p)));}
    void Upd(int t,int x,int y,int z){for(;x<=lim;x+=x&-x)upd(rt[t][x],1,lim,y,z);}int Qry(int t,int x,int y){int res=-1e9;for(;x;x^=x&-x)res=max(res,qry(rt[t][x],1,lim,y));return res;}
    void main()//q->R k->B
    {
        ios::sync_with_stdio(0),cin>>n>>q>>k>>(ch+1),m=q+k,lim=n<<1,v[0]=-1e9;
        for(int i=1;i<=n;++i)a[i]=a[i-1]+(ch[i]=='R'),b[i]=b[i-1]+(ch[i]=='B');a[0]=b[0]=n+1;for(int i=1;i<=n;++i)x=i/m,a[i]=n+1+x*q-a[i],b[i]=n+1+x*k-b[i]/*,assert(1<=min(a[i],b[i])&&max(a[i],b[i])<=lim),printf("%d %d\n",a[i],b[i])*/;
        for(int i=0;i<=n;++i)x=i%m+1,y=i/m,f[i]=(i?max(f[i-1],Qry(x,a[i],b[i])+y):0),Upd(x,a[i],b[i],f[i]-y)/*,printf("%d%c",f[i]," \n"[!((i+1)%m)])*/;
        // for(int i=0;i<=n;i+=4)printf("%d %d\n",a[i],b[i]);
        cout<<f[n]<<endl;
    }
};
int main()
{
    const bool base=0,IO=1;int T;if(!base)T=1;else if(IO)T=read();else ios::sync_with_stdio(0),cin>>T;for(;T--;)Main::main();
    return 0;
}