#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#define eps 1e-9

using namespace std;

const int maxn=100+5;

const int INF=1e5;

int n;

char s[maxn];

double cosA[maxn];

double p[maxn*maxn];

double ans[maxn*maxn];

struct Facility

{

    char c;

    int x;

    int y;

}fac[maxn];

struct Angle

{

    char c;

    double cosA;

}ang[maxn];

bool cmp(Angle a,Angle b)

{

    return a.cosA<b.cosA;

}

bool judge(double o)

{

    for(int i=0;i<n;i++)

    {

        ang[i].c=fac[i].c;

        ang[i].cosA=(fac[i].x-o)/sqrt(fac[i].y*fac[i].y+(fac[i].x-o)*(fac[i].x-o)); //算区间中点与建筑的连线与X轴正方向的夹角的余弦值

    }

    sort(ang,ang+n,cmp);

    for(int i=0;i<n;i++)

        if(ang[i].c!=s[i])

            return false;

    return true;

}

int main()

{

//    freopen("in.txt","r",stdin);

//    freopen("area.in","r",stdin);

//    freopen("area.out","w",stdout);

    while(~scanf("%d",&n))

    {

        getchar();

        scanf("%s",s);

        for(int i=0;i<n;i++)

        {

            getchar();

            scanf("%c%d%d",&fac[i].c,&fac[i].x,&fac[i].y);

        }

        p[0]=-INF;

        int cnt=1;

        for(int i=0;i<n;i++)

        {

            for(int j=i+1;j<n;j++)

            {

                if(fac[i].y==fac[j].y) continue;

                p[cnt++]=1.0*(fac[j].x*fac[i].y-fac[i].x*fac[j].y)/(fac[i].y-fac[j].y);

            }

        }

        p[cnt++]=INF;

        sort(p,p+cnt);

        int cnt2=0;

        for(int i=0;i<cnt-1;i++)

        {

            if(p[i+1]-p[i]<eps) continue;

            if(judge((p[i]+p[i+1])/2))  //去区间中点来判断此区间是否合法

            {

                ans[cnt2++]=p[i];

                ans[cnt2++]=p[i+1];

            }

        }

        printf("%d\n",cnt2/2);

        if(cnt2==0) continue;

        if(fabs(ans[0]-p[0])<eps)

            printf("%c ",'*');

        else

            printf("%.7f ",ans[0]);

        for(int i=1;i<cnt2-1;i++)

            printf("%.7f ",ans[i]);

        if(fabs(ans[cnt2-1]-p[cnt-1])<eps)

            printf("%c\n",'*');

        else

            printf("%.7f\n",ans[cnt2-1]);

    }

    return 0;

}