import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;

/**
 * @author icewei
 * @Description: 双链表  acwing-ycx
 * @date 2025/1/26
 */
public class Main {

    private static int N = 100000 + 5;
    private static int[] e = new int[N];
    private static int[] l = new int[N];
    private static int[] r = new int[N];
    private static int idx = 0;

    private static void init() {
        // 0表示左端点，1表示右端点
        r[0] = 1; l[1] = 0; idx = 2;
    }

    public static void main(String[] args) throws IOException {
        init();

        StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        st.nextToken(); int m = (int)st.nval;
        for (int i = 0; i < m; i++) {
            st.nextToken(); String op = st.sval;
            if ("L".equals(op)) {
                // 在最左端插入 x
                st.nextToken(); int x = (int)st.nval;
                add(0, x);
            } else if ("R".equals(op)) {
                // 在最右端插入 x
                st.nextToken(); int x = (int)st.nval;
                addLeft(1, x);
            } else if ("D".equals(op)) {
                // 删除第 k 个插入的数，也就是删除下标是 k+1的数
                st.nextToken(); int k = (int)st.nval;
                del(k);
            } else if ("IL".equals(op)) {
                st.nextToken(); int k = (int)st.nval;
                st.nextToken(); int x = (int)st.nval;
                addLeft(k + 1, x);
            } else if ("IR".equals(op)) {
                st.nextToken(); int k = (int)st.nval;
                st.nextToken(); int x = (int)st.nval;
                add(k + 1, x);
            }
        }

        for(int i = 0; i != 1; i = r[i]) {
            if (i != 0) System.out.print(e[i] + " ");
        }
        System.out.println();
    }

    // 在下表是 k 的点的右边，插入一个点
    private static void add(int k, int x) {
        e[idx] = x; r[idx] = r[k]; l[idx] = k; l[r[k]] = idx; r[k] = idx; idx++;
    }

    // 在下表是 k 的点的左边，插入一个点
    private static void addLeft(int k, int x) {
        add(l[k], x);
    }

    // 删除第 k 个插入的数，下标是 k+1
    private static void del(int k) {
        r[l[k + 1]] = r[k + 1];
        l[r[k+1]] = l[k + 1];
    }
}