In some country there are exactly n cities and m bidirectional roads connecting the cities. Cities are numbered with integers from 1 to n. If cities a and b are connected by a road, then in an hour you can go along this road either from city a to city b, or from city b to city a. The road network is such that from any city you can get to any other one by moving along the roads.

You want to destroy the largest possible number of roads in the country so that the remaining roads would allow you to get from city s_1 to city t_1 in at most l_1 hours and get from city s_2 to city t_2 in at most l_2 hours.

Determine what maximum number of roads you need to destroy in order to meet the condition of your plan. If it is impossible to reach the desired result, print -1.


-----Input-----

The first line contains two integers n, m (1 ≤ n ≤ 3000, $n - 1 \leq m \leq \operatorname{min} \{3000, \frac{n(n - 1)}{2} \}$) — the number of cities and roads in the country, respectively. 

Next m lines contain the descriptions of the roads as pairs of integers a_{i}, b_{i} (1 ≤ a_{i}, b_{i} ≤ n, a_{i} ≠ b_{i}). It is guaranteed that the roads that are given in the description can transport you from any city to any other one. It is guaranteed that each pair of cities has at most one road between them.

The last two lines contains three integers each, s_1, t_1, l_1 and s_2, t_2, l_2, respectively (1 ≤ s_{i}, t_{i} ≤ n, 0 ≤ l_{i} ≤ n).


-----Output-----

Print a single number — the answer to the problem. If the it is impossible to meet the conditions, print -1.


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

Output
0

Input
5 4
1 2
2 3
3 4
4 5
1 3 2
2 4 2

Output
1

Input
5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 1

Output
-1

Difficulty: competition

Starter code:

Solutions:
from itertools import combinations_with_replacement 
from collections import deque

#sys.stdin = open("input_py.txt","r")

n, m = map(int, input().split())
G = [ [] for i in range(n)]

for i in range(m):
    x, y = map(int, input().split())
    x-=1; y-=1
    G[x].append(y)
    G[y].append(x)

def BFS(s):
    dist = [-1 for i in range(n)]
    dist[s] = 0
    Q = deque()
    Q.append(s)
    while len(Q) > 0:
        v = Q.popleft()
        for to in G[v]:
            if dist[to] < 0:
                dist[to] = dist[v] + 1
                Q.append(to)
    return dist 


Dist = [BFS(i) for i in range(n)]

s1, t1, l1 = map(int, input(). split())
s2, t2, l2 = map(int, input(). split())
s1-=1; t1-=1; s2-=1; t2-=1
if Dist[s1][t1] > l1 or Dist[s2][t2] > l2:
    print(-1)
    return

rest = Dist[s1][t1] + Dist[s2][t2]

for i in range(n):
    for j in range(n):
        if Dist[i][s1] + Dist[i][j] + Dist[j][t1] <= l1 and Dist[i][s2] + Dist[i][j] + Dist[j][t2] <= l2 :
            rest = min(rest, Dist[i][j] + Dist[i][s1] + Dist[i][s2] + Dist[j][t1] + Dist[j][t2])
        if Dist[i][s1] + Dist[i][j] + Dist[j][t1] <= l1 and Dist[j][s2] + Dist[i][j] + Dist[i][t2] <= l2 :
            rest = min(rest, Dist[i][j] + Dist[j][t1] + Dist[j][s2] + Dist[i][s1] + Dist[i][t2])
print(m-rest)

import sys
from collections import deque

n, m = [int(i) for i in sys.stdin.readline().split()]
neighbors = [set() for _ in range(n)]
for i in range(m):
    m1, m2 = [int(i) for i in sys.stdin.readline().split()]
    neighbors[m1-1].add(m2-1)
    neighbors[m2-1].add(m1-1)

s1, t1, l1 = [int(i) for i in sys.stdin.readline().split()]
s2, t2, l2 = [int(i) for i in sys.stdin.readline().split()]
s1 -= 1
s2 -= 1
t1 -= 1
t2 -= 1

# compute all pairs distance
dists = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n):
    # bfs
    q = deque([(i, 0)])
    visited = [False for _ in range(n)]
    visited[i] = True
    while len(q) != 0:
        v, dist = q.popleft()
        dists[i][v] = dist
        for neighbor in neighbors[v]:
            if not visited[neighbor]:
                q.append((neighbor, dist+1))
                visited[neighbor] = True

best_found = m+1
# first case: 2 paths don't share edge
if dists[s1][t1] <= l1 and dists[s2][t2] <= l2:
    best_found = min(best_found, dists[s1][t1] + dists[s2][t2])

# second case: there are points u, v such that they share path along u-v
for u in range(n):
    for v in range(n):
        if u == v:
            continue
        # case 1: s1-u-v-t1, s2-u-v-t2
        path1_length = dists[s1][u] + dists[u][v] + dists[v][t1]
        path2_length = dists[s2][u] + dists[u][v] + dists[v][t2]
        if path1_length <= l1 and path2_length <= l2:
            total_length = path1_length + path2_length - dists[u][v]
            best_found = min(best_found, total_length)
        # case 2: s1-u-v-t1, s2-v-u-t2
        path1_length = dists[s1][u] + dists[u][v] + dists[v][t1]
        path2_length = dists[s2][v] + dists[v][u] + dists[u][t2]
        if path1_length <= l1 and path2_length <= l2:
            total_length = path1_length + path2_length - dists[u][v]
            best_found = min(best_found, total_length)
        # case 3: s1-v-u-t1, s2-u-v-t2
        path1_length = dists[s1][v] + dists[v][u] + dists[u][t1]
        path2_length = dists[s2][u] + dists[u][v] + dists[v][t2]
        if path1_length <= l1 and path2_length <= l2:
            total_length = path1_length + path2_length - dists[u][v]
            best_found = min(best_found, total_length)
        # case 4: s1-v-u-t1, s2-v-u-t2
        path1_length = dists[s1][v] + dists[v][u] + dists[u][t1]
        path2_length = dists[s2][v] + dists[v][u] + dists[u][t2]
        if path1_length <= l1 and path2_length <= l2:
            total_length = path1_length + path2_length - dists[u][v]
            best_found = min(best_found, total_length)

print(m - best_found)