human
I found an interesting problem on https://codeforces.com/problemset/problem/546/E:
In the country there are n cities and m bidirectional roads between them. Each city has an army. Army of the i-th city consists of a_{i} soldiers. Now soldiers roam. After roaming each soldier has to either stay in his city or to go to the one of neighboring cities by at moving along at most one road.

Check if is it possible that after roaming there will be exactly b_{i} soldiers in the i-th city.

-----Input-----

First line of input consists of two integers n and m (1 ≤ n ≤ 100, 0 ≤ m ≤ 200).

Next line contains n integers a_1, a_2, ..., a_{n} (0 ≤ a_{i} ≤ 100).

Next line contains n integers b_1, b_2, ..., b_{n} (0 ≤ b_{i} ≤ 100).

Then m lines follow, each of them consists of two integers p and q (1 ≤ p, q ≤ n, p ≠ q) denoting that there is an undirected road between cities p and q. 

It is guaranteed that there is at most one road between each pair of cities.

-----Output-----

If the conditions can not be met output single word "NO".

Otherwise output word "YES" and then n lines, each of them consisting of n integers. Number in the i-th line in the j-th column should denote how many soldiers should road from city i to city j (if i ≠ j) or how many soldiers should stay in city i (if i = j).

If there are several possible answers you may output any of them.

-----Examples-----
Input
4 4
1 2 6 3
3 5 3 1
1 2
2 3
3 4
4 2

Output
YES
1 0 0 0 
2 0 0 0 
0 5 1 0 
0 0 2 1 

Input
2 0
1 2
2 1

Output
NO

I tried it in Python, but could not do it. Can you solve it?

 ... Write me a the Python program you requested?
assistant
#Here is the Python program you requested.

from collections import defaultdict
import sys

N, M = input().split(' ')
N, M = int(N), int(M)
A = input().split(' ')
A = list(map(int, A))
B = input().split(' ')
B = list(map(int, B))

if (sum(A) != sum(B)):
    print('NO')
    return

neighbours = defaultdict(lambda: defaultdict(int))
flow = defaultdict(int)

def left(i):
    return 'l ' + str(i)

def right(i):
    return 'r ' + str(i)

for i in range(1, N+1):
    for j in range(1, N+1):
        flow[left(i), right(j)] = 0
        flow[right(i), left(j)] = 0

for i, capacity in enumerate(A):
    neighbours['source'][left(i + 1)] = capacity
for i, capacity in enumerate(B):
    neighbours[right(i + 1)]['sink'] = capacity
for i in range(1, N+1):
    neighbours[left(i)][right(i)] = float('inf')

for m in range(M):
    p, q = input().split(' ')
    p, q = int(p), int(q)
    neighbours[left(p)][right(q)] = float('inf')
    neighbours[left(q)][right(p)] = float('inf')

def search():
    visited = set()
    def dfs(v):
        if v in visited:
            return False
        if v == 'sink':
            return ['sink']
        visited.add(v)
        for n in neighbours[v]:
            capacity = neighbours[v][n]
            if capacity > 0:
                res = dfs(n)
                if res:
                    return [v] + res
    return dfs('source')

total_flow = 0

def augment_edge(a, b, amount):
    neighbours[a][b] -= amount
    neighbours[b][a] += amount
    flow[a, b] += amount

def augment(path):
    capacity = float('inf')
    for i in range(len(path) - 1):
        capacity = min(capacity, neighbours[path[i]][path[i+1]])
    # print('augment', path, capacity)
    for i in range(len(path) - 1):
        augment_edge(path[i], path[i+1], capacity)
    nonlocal total_flow
    total_flow += capacity

while True:
    res = search()
    if res is None:
        break
    else:
        augment(res)

if (total_flow == sum(A)):
    print("YES")
    for i in range(1, N+1):
        for j in range(1, N+1):
            print(flow[left(i), right(j)] - flow[right(j),...