Codeforces Round #199 (Div. 2) -- B. Xenia and Spies （简单模拟）

B. Xenia and Spies

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 ton from left to right.

Spy s has an important note. He has to pass the note to spyf. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can pass the note to one of his neighbours in the row. In other words, if this spy's number isx, he can pass the note to another spy, eitherx - 1 or x + 1 (ifx = 1 or x = n, then the spy has only one neighbour). Also during a step the spy can keep a note and not pass it to anyone.

But nothing is that easy. During m steps Xenia watches some spies attentively. Specifically, during stept_i (steps are numbered from 1) Xenia watches spies numbersl_i, l_i + 1, l_i + 2, ..., r_i(1 ≤ l_i ≤ r_i ≤ n). Of course, if during some step a spy is watched, he can't do anything: neither give the note nor take it from some other spy. Otherwise, Xenia reveals the spies' cunning plot. Nevertheless, if the spy at the current step keeps the note, Xenia sees nothing suspicious even if she watches him.

You've got s and f. Also, you have the steps during which Xenia watches spies and which spies she is going to watch during each step. Find the best way the spies should act in order to pass the note from spys to spy f as quickly as possible (in the minimum number of steps).

Input

The first line contains four integers n,m, s andf (1 ≤ n, m ≤ 10⁵; 1 ≤ s, f ≤ n; s ≠ f; n ≥ 2). Each of the followingm lines contains three integers t_i, l_i, r_i(1 ≤ t_i ≤ 10⁹, 1 ≤ l_i ≤ r_i ≤ n). It is guaranteed that t₁ < t₂ < t₃ < ... < t_m.

Output

Print k characters in a line: the i-th character in the line must represent the spies' actions on stepi. If on step i the spy with the note must pass the note to the spy with a lesser number, thei-th character should equal "L". If on stepi the spy with the note must pass it to the spy with a larger number, thei-th character must equal "R". If the spy must keep the note at thei-th step, the i-th character must equal "X".

As a result of applying the printed sequence of actions spy s must pass the note to spy f. The number of printed charactersk must be as small as possible. Xenia must not catch the spies passing the note.

If there are miltiple optimal solutions, you can print any of them. It is guaranteed that the answer exists.

Examples

Input

3 5 1 31 1 22 2 33 3 34 1 110 1 3

Output

XXRR大体题意：有n 个间谍，标号为1到n，有m 行检查， 告诉你起点标号s，终点标号f。问如何去走才能从起点到终点！思路：简单模拟即可！当s！= f 就一直循环。先算出下一步是L 还是R，然后看看是不是在检查区间内，在的话 +X，否则 加ch = L || R！注意，连续的两个检查 时间可能跳度非常大，注意时间不想等就一直走就好了！
#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;const int maxn = 1000000 + 10;vector<char>ans;char st[maxn];int l[maxn],r[maxn],t[maxn];int main(){    int n,m,s,f;    while(scanf("%d %d %d %d",&n,&m,&s,&f) == 4 ){        for (int i = 0; i < m ;++i){            scanf("%d %d %d",&t[i],&l[i],&r[i]);        }        int cnt = 0;        int i=0,j=1;        while(s != f){//        printf("%d\n",cnt);            int ns = s;            int ok = 1;            char ch;            if (s < f){ns = s+1;ch = 'R';}            else {                ns = s-1;                ch = 'L';            }            if (i < m && j == t[i]){                if (s >= l[i] && s <= r[i] || ns >= l[i] && ns <= r[i])ok=0;                i++;            }            if (ok)st[cnt++]=ch,s=ns;            else st[cnt++] = 'X';            ++j;        }        st[cnt] = 0;        puts(st);    }    return 0;}