#include <iostream>
#include <cstdio>
#include <map>
#define mid ((l + r) >> 1)
using namespace std;

const int inf = 1e9;

int n,R,B;
char s[200010];
int sumb[200010],sumr[200010],bmin,bmax,rmin,rmax;
struct node2
{
    int ls,rs,val;
} tree[400010 * 350];
struct node
{
    int ls,rs;
} tree1[400010 * 23];
int tcnt,Rt[400010],ttcnt,tcnt1;
map <int,int> rt;
int f[200010];

void upd2(int &x,int l,int r,int qx,int k)
{
    if (!x) x = ++ tcnt,tree[x].val = -inf;
    tree[x].val = max(tree[x].val,k);
    if (l == r) return ;
    if (qx <= mid) upd2(tree[x].ls,l,mid,qx,k);
    else upd2(tree[x].rs,mid + 1,r,qx,k);
}

void upd1(int &x,int l,int r,int qx,int qy,int k)
{
    if (!x) x = ++ tcnt1;
    upd2(rt[x],rmin,rmax,qy,k);
    if (l == r) return ;
    if (qx <= mid) upd1(tree1[x].ls,l,mid,qx,qy,k);
    else upd1(tree1[x].rs,mid + 1,r,qx,qy,k);
}

int query2(int x,int l,int r,int qx)
{
    if (!x || r < qx) return -inf;
    if (l >= qx) return tree[x].val;
    return max(query2(tree[x].ls,l,mid,qx),query2(tree[x].rs,mid + 1,r,qx));
}

int query1(int x,int l,int r,int qx,int qy)
{
    if (!x || r < qx) return -inf;
    if (l >= qx) return query2(rt[x],rmin,rmax,qy);
    return max(query1(tree1[x].ls,l,mid,qx,qy),query1(tree1[x].rs,mid + 1,r,qx,qy));
}

int main()
{
    scanf("%d%d%d",&n,&R,&B);
    scanf("%s",s + 1);
    for (int i = 1;i <= n;i ++)
    {
        sumb[i] = sumb[i - 1] + (s[i] == 'B');
        sumr[i] = sumr[i - 1] + (s[i] == 'R');
        int k = i / (R + B);
        bmin = min(bmin,sumb[i] - k * B),bmax = max(bmax,sumb[i] - k * B);
        rmin = min(rmin,sumr[i] - k * R),rmax = max(rmax,sumr[i] - k * R);
    }
    upd1(Rt[0],bmin,bmax,0,0,0);
    for (int i = 1;i <= n;i ++)
    {
        int opt = i % (R + B),k = i / (R + B);
        int tmp = query1(Rt[opt],bmin,bmax,sumb[i] - k * B,sumr[i] - k * R);
        f[i] = (tmp >= -inf ? tmp + k : 0);
        f[i] = max(f[i],f[i - 1]);
        // printf("%d %d %d\n",i,f[i],tmp);
        upd1(Rt[opt],bmin,bmax,sumb[i] - k * B,sumr[i] - k * R,f[i] - k);
        // printf("upd %d %d %d %d\n",i,sumb[i] - k * B,sumr[i] - k * R,f[i] - k);
        // printf("*%d\n",query1(Rt[0],-2e5,2e5,-1,1));
    }
    // if (n == 199998 && R == 66667 && B == 66665) printf("%d\n",tcnt);
    printf("%d\n",f[n]);
}