inputs: I found an interesting problem on https://codeforces.com/problemset/problem/733/C:
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 a_{i}.

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:  the first monster can't eat the second monster because a_1 = 1 is not greater than a_2 = 2;  the second monster can't eat the third monster because a_2 = 2 is not greater than a_3 = 2;  the second monster can't eat the fifth monster because they are not neighbors;  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 b_{j}. 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...

I tried it in Python, but could not do it. Can you solve it?
targets: import sys

a = [0,]
b = [0,]
ans1 = []
ans2 = []
n = int(input())
s = input()
nums = s.split()
for i in range(0, n):
    a.append(int(nums[i]))

k = int(input())
s = input()
nums = s.split()
for i in range(0, k):
    b.append(int(nums[i]))

def f(x, y, z):
    #print(x,y,z)
    pos1 = x
    pos2 = x
    if x == y:
        return 1
    for i in range(x, y + 1):
        if a[i] > a[pos1]:
            pos1 = i
        if a[i] >= a[pos2]:
            pos2 = i
    for i in range(x, y):
        if a[i] == a[pos2]:
            if a[i + 1] < a[i]:
                pos2 = i
    for i in range(x + 1, y + 1):
        if a[i] == a[pos1]:
            if a[i - 1] < a[i]:
                pos1 = i
    if pos1 != x or a[pos1] > a[pos1 + 1]:
        for i in range(0, pos1 - x):
            ans1.append(pos1 - x + z - i)
            ans2.append('L')
        for i in range(0, y - pos1):
            ans1.append(z)
            ans2.append('R')
    elif pos2 != y or a[pos2] > a[pos2 - 1]:
        for i in range(0, y - pos2):
            ans1.append(pos2 - x + z)
            ans2.append('R')
        for i in range(0, pos2 - x):
            ans1.append(pos2 - x + z - i)
            ans2.append('L')
    else:
        return 0

    return 1

lasti = 0
j = 1
sum = 0
for i in range(1, n+1):
    if j > k:
        print('NO')
        return
    sum += a[i]
    #print(i, sum, j)
    if sum > b[j]:
        print('NO')
        return
    if sum == b[j]:
        if f(lasti + 1, i, j) == 0:
            print('NO')
            return
        lasti = i
        j += 1
        sum = 0

if j <= k:
    print('NO')
    return

print('YES')
for i in range(0, len(ans1)):
    print(ans1[i], ans2[i])