#include <bits/stdc++.h> // By Lucky Ox
#define int long long
#define endl "\n"
#define pii pair<int, int>
#define PI atan(1.0) * 4
using namespace std;
using i128 = __int128;
typedef unsigned long long ull;
const long long mod = 1000000000000002493, INF = 0x3f3f3f3f3f3f3f3f;
int P(int x, int p){ return (x % p + p) % p; }
int lcm(int x, int y) { return x / gcd(x, y) * y; }
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int len_10(int x) { int len = 0; while(x) { x /= 10; len ++ ; } return len; }
int q_pow(int a, int k, int p) { int res = 1; while (k) { if (k & 1) res = res * a % p; k >>= 1; a = a * a % p; } return res; }
int to_int(string s) { int val = 0; for(int i = 0; i < (int)s.size(); i ++ ){val *= 10; val += s[i] - '0';}return val;}//注意:s是空串也会返回0
i128 read() { i128 x = 0; char c = getchar(); while (c < '0' || c > '9') c = getchar(); while (c >= '0' && c <= '9') {
x = x * 10 + c - '0'; c = getchar(); } return x; } //i128输入
void print(i128 x) {if(x > 9) print(x / 10); putchar(x % 10 + '0'); }//i128输出
//用__lg()来求一个数二进制下的位数 返回的len 表示这个数是[0, 1, ...., len] 比如10 __lg(10) = 3, 1010 [3, 2, 1, 0]
//__builtin_popcountll(int x) 求二进制下x中1的数量 __buitlin_ctzll(int x) 求二进制下末尾0的个数
//(n & (1 << i))的值可能会是1, 2, 4...... (n >> i) & 1的值一定是1
//将x转换成[1, n]中对应的数 (x - 1) % n + 1

const int N = 2e3 + 10;
int n;
struct {
    string s;
    int x, y, r;
    int x1, y1, x2, y2;
    char col;
}a[N];

struct {
    int x1, y1, x2, y2;
}c[N];

int f(int x, int y, int cj) {
    int val = (x - a[cj].x) * (x - a[cj].x) + (y - a[cj].y) * (y - a[cj].y);
    if(val <= a[cj].r * a[cj].r) return 1;
    return 0; 
}

int F(int x, int y, int cj) {
    if(x >= a[cj].x1 && x <= a[cj].x2 && y >= a[cj].y1 && y <= a[cj].y2) return 1;
    return 0;
}

void solve() {
    cin >> n;

    int lena = 0, lenc = 0;
    for(int i = 1; i <= n; i ++ ) {
        string s;
        cin >> s;

        if(s == "Circle") {
            int x, y, r;
            char col;
            cin >> x >> y >> r >> col;

            lena ++ ;
            a[lena].x = x; a[lena].y = y; a[lena].r = r; 
            a[lena].s = s; a[lena].col = col;
        }
        else if(s == "Rectangle") {
            int x1, y1, x2, y2;
            char col;
            cin >> x1 >> y1 >> x2 >> y2 >> col;

            lena ++ ;
            a[lena].x1 = x1, a[lena].y1 = y1, a[lena].x2 = x2, a[lena].y2 = y2;
            a[lena].s = s; a[lena].col = col;
        }
        else {
            int x1, y1, x2, y2;
            cin >> x1 >> y1 >> x2 >> y2;

            c[++ lenc] = {x1, y1, x2, y2};
        }
    }

    for(int i = 1; i <= lenc; i ++ ) {
        int x1 = c[i].x1, y1 = c[i].y1, x2 = c[i].x2, y2 = c[i].y2;

        for(int y = y2; y >= y1; y -- ) {
             for(int x = x1; x <= x2; x ++ ) {
                char ans = '.';

                for(int j = lena; j >= 1; j -- ) {
                    if(a[j].s == "Circle") {
                        if(f(x, y, j)) {
                            ans = a[j].col;
                            break;
                        }
                    }
                    else if(a[j].s == "Rectangle") {
                        if(F(x, y, j)) {
                            ans = a[j].col;
                            break;
                        }
                    }
                }
                cout << ans;
            }
            cout << endl;
        }
    }
}

signed main() {
    ios::sync_with_stdio(0);cin.tie(0);
    cout << fixed << setprecision(10);
    int T = 1;
    //cin >> T;
    while (T -- ) solve();
    return 0;
}
/*
6
Circle 0 0 5 *
Circle -2 2 1 @
Circle 2 2 1 @
Rectangle 0 -1 0 0 ^
Rectangle -2 -2 2 -2 _
Render -5 -5 5 5
*/