Codeforces Round #302 (Div. 1)B. Destroying Roads 最短路

B. Destroying Roads

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 s₁ to city t₁ in at most l₁ hours and get from city s₂ to city t₂ in at most l₂ 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, ) — 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, s₁, t₁, l₁ and s₂, t₂, l₂, 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.

Sample test(s)

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

题意：n个城市，m条边，现在想删除尽量多的边，使得s1(s2)->t1(t2)的路径长度最多为l1(l2)

思路：这个题主要是两条路径重复的部分的计算，可以用枚举的方法暴力计算。首先bfs暴力求出任意两个点之间的最短路，然后暴力枚举两条路重复的部分，取最小值，注意枚举的时候，要两个方向都枚举

#include
using namespace std;
const int INF=1000000000;
const int maxn=3101;
vector g[maxn];
int N,M;
int s1,t1,l1,s2,t2,l2;
int dis[maxn][maxn],vis[maxn];
void BFS(int st)
{
    queue q;
    for(int i=1;i<=N;i++)
        dis[st][i]=INF,vis[i]=0;
    dis[st][st]=0;
    q.push(st);
    while(!q.empty())
    {
        int u=q.front();q.pop();
        int len=g[u].size();
        vis[u]=1;
        for(int i=0;il1||dis[s2][t2]>l2)
    {
        printf("-1\n");
        return 0;
    }
    int ans=dis[s1][t1]+dis[s2][t2];
    for(int i=1;i<=N;i++)
    {
        for(int j=1;j<=N;j++)
        {
            int ll1=dis[s1][i]+dis[i][j]+dis[j][t1];
            int ll2=dis[s1][j]+dis[j][i]+dis[i][t1];
            int ll3=dis[s2][i]+dis[i][j]+dis[j][t2];
            if(ll1<=l1&&ll3<=l2)ans=min(ans,ll1+ll3-dis[i][j]);
            if(ll2<=l1&&ll3<=l2)ans=min(ans,ll2+ll3-dis[i][j]);
        }
    }
    printf("%d\n",M-ans);
    return 0;
}