Codeforces Round #378 (Div. 2)-C. Epidemic in Monstropolis

原题链接

C. Epidemic in Monstropolis

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There was an epidemic in Monstropolis and all monsters became sick. To recover, all monsters lined up in queue for an appointment to the only doctor in the city.

Soon, monsters became hungry and began to eat each other.

One monster can eat other monster if its weight is strictly greater than the weight of the monster being eaten, and they stand in the queue next to each other. Monsters eat each other instantly. There are no monsters which are being eaten at the same moment. After the monster A eats the monster B, the weight of the monster A increases by the weight of the eaten monster B. In result of such eating the length of the queue decreases by one, all monsters after the eaten one step forward so that there is no empty places in the queue again. A monster can eat several monsters one after another. Initially there were n monsters in the queue, the i-th of which had weight ai.

For example, if weights are [1, 2, 2, 2, 1, 2] (in order of queue, monsters are numbered from 1 to 6 from left to right) then some of the options are:

1. the first monster can't eat the second monster because a1 = 1 is not greater than a2 = 2;
2. the second monster can't eat the third monster because a2 = 2 is not greater than a3 = 2;
3. the second monster can't eat the fifth monster because they are not neighbors;
4. the second monster can eat the first monster, the queue will be transformed to [3, 2, 2, 1, 2].

After some time, someone said a good joke and all monsters recovered. At that moment there were k (k ≤ n) monsters in the queue, the j-th of which had weight bj. Both sequences (a and b) contain the weights of the monsters in the order from the first to the last.

You are required to provide one of the possible orders of eating monsters which led to the current queue, or to determine that this could not happen. Assume that the doctor didn't make any appointments while monsters were eating each other.

Input

The first line contains single integer n (1 ≤ n ≤ 500) — the number of monsters in the initial queue.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the initial weights of the monsters.

The third line contains single integer k (1 ≤ k ≤ n) — the number of monsters in the queue after the joke.

The fourth line contains k integers b1, b2, ..., bk (1 ≤ bj ≤ 5·108) — the weights of the monsters after the joke.

Monsters are listed in the order from the beginning of the queue to the end.

Output

In case if no actions could lead to the final queue, print "NO" (without quotes) in the only line.

Otherwise print "YES" (without quotes) in the first line. In the next n - k lines print actions in the chronological order. In each line print x — the index number of the monster in the current queue which eats and, separated by space, the symbol 'L' if the monster which stays the x-th in the queue eats the monster in front of him, or 'R' if the monster which stays the x-th in the queue eats the monster behind him. After each eating the queue is enumerated again.

When one monster eats another the queue decreases. If there are several answers, print any of them.

Examples

input

61 2 2 2 1 225 5

output

YES2 L1 R4 L3 L

input

51 2 3 4 5115

output

YES5 L4 L3 L2 L

input

51 1 1 3 332 1 6

output

NO

a数组的某个后缀a[l],..a[r]和等于b数组的最后一个数,每次取l..r中重量最大的怪兽，吃掉旁边的怪兽，观察是否最终只剩一个怪兽.按照这个过程不断进行

注意，不仅要观察b数组个怪兽是否全部由a形成，还要判断a数组的怪兽最后是否剩余

#include <bits/stdc++.h>#define maxn 505#define MOD 1000000007using namespace std;typedef long long ll;ll p1[maxn], p2[maxn];vector<int> v1, v2;int main(){//freopen("in.txt", "r", stdin);int n, m, len;scanf("%d", &n);for(int i = 1; i <= n; i++) scanf("%I64d", p1+i);scanf("%d", &m);for(int i = 1; i <= m; i++) scanf("%I64d", p2+i);len = n;for(int i = m; i >= 1; i--){int j;ll e = 0;for(j = len; j >= 1; j--){e += p1[j];if(e >= p2[i]) break;}if(e != p2[i]){puts("NO");return 0;}int l = j, r = len;len = j - 1;while(l < r){ll s = 0;int k, d1 = 0, d2 = 0;for(int h = l; h <= r; h++){if(h - 1 >= l && p1[h] > p1[h-1]){if(p1[h] > s){k = h;s = p1[h];d1 = 1;d2 = 0;}}if(h + 1 <= r && p1[h] > p1[h+1]){if(p1[h] > s){k = h;s = p1[h];d2 = 1;d1 = 0;}}}if(s == 0){puts("NO");return 0;}int mm;if(d1){p1[k] += p1[k-1];mm = k;v1.push_back(k);v2.push_back('L');}if(d2){p1[k] += p1[k+1];mm = k + 2;v1.push_back(k);v2.push_back('R');}for(int q = mm; q <= r; q++){   p1[q-1] = p1[q]; }r--;}}if(len != 0){puts("NO");return 0;}puts("YES");for(int i = 0; i < v1.size(); i++){printf("%d %c\n", v1[i], v2[i]);}return 0;}