CF804D：Expected diameter of a tree（树的直径 & dfs）

D. Expected diameter of a tree

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Pasha is a good student and one of MoJaK's best friends. He always have a problem to think about. Today they had a talk about the following problem.

We have a forest (acyclic undirected graph) with n vertices and m edges. There are q queries we should answer. In each query two vertices v and u are given. Let V be the set of vertices in the connected component of the graph that contains v, and U be the set of vertices in the connected component of the graph that contains u. Let's add an edge between some vertex  and some vertex in  and compute the value d of the resulting component. If the resulting component is a tree, the value d is the diameter of the component, and it is equal to -1 otherwise. What is the expected value of d, if we choose vertices a and b from the sets uniformly at random?

Can you help Pasha to solve this problem?

The diameter of the component is the maximum distance among some pair of vertices in the component. The distance between two vertices is the minimum number of edges on some path between the two vertices.

Note that queries don't add edges to the initial forest.

Input

The first line contains three integers n, m and q(1 ≤ n, m, q ≤ 105) — the number of vertices, the number of edges in the graph and the number of queries.

Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n), that means there is an edge between vertices ui and vi.

It is guaranteed that the given graph is a forest.

Each of the next q lines contains two integers ui and vi (1 ≤ ui, vi ≤ n) — the vertices given in the i-th query.

Output

For each query print the expected value of d as described in the problem statement.

Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6. Let's assume that your answer is a, and the jury's answer is b. The checker program will consider your answer correct, if .

Examples

input

3 1 2
1 3
3 1
2 3

output

-1
2.0000000000

input

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

output

-1
2.6666666667
2.6666666667

Note

In the first example the vertices 1 and 3 are in the same component, so the answer for the first query is -1. For the second query there are two options to add the edge: one option is to add the edge 1 - 2, the other one is 2 - 3. In both ways the resulting diameter is 2, so the answer is 2.

In the second example the answer for the first query is obviously -1. The answer for the second query is the average of three cases: for added edges 1 - 2 or 1 - 3 the diameter is 3, and for added edge 1 - 4 the diameter is 2. Thus, the answer is .

题意：给一片森林，q个询问，每个询问两个点，问将这两个点所在的集合连接起来组成的新集合，它的最远两点的距离的期望值是多少。

思路：先预处理每棵树的直径z[]及里面的每个点离最远点的距离dis[i]，然后枚举A树的点i，二分B树的点j，二分这个dis[i]+dis[j]+1 > max(z[A], z[B])。

# include <bits/stdc++.h>
# define pb push_back
# define mp make_pair
# define PII pair<int, int>
# define A first
# define B second
# define LL long long
using namespace std;
const int maxn = 1e5+8;
int n, dis[maxn]={0}, f[maxn]={0}, z[maxn]={0};
vector<int>v[maxn], son[maxn],str[maxn];
vector<LL>pre[maxn];
map<PII, double>M;
PII dfs1(int cur, int pre, int root)
{
    f[cur] = root;
    PII now = mp(0, cur);
    for(auto to : v[cur])
    {
        if(to == pre) continue;
        PII tmp = dfs1(to, cur, root);
        if(tmp.A +1 > now.A) now = mp(tmp.A+1, tmp.B);
    }
    return now;
}
void dfs2(int cur, int pre, int d)
{
    dis[cur] = max(dis[cur], d);
    for(auto to : v[cur])
    {
        if(to == pre) continue;
        dfs2(to, cur, d+1);
    }
}
void init()
{
    for(int i=1; i<=n; ++i)
    {
        if(f[i] == 0)
        {
            PII a = dfs1(i, 0, i);
            PII b = dfs1(a.B, 0, i);
            dfs2(a.B, 0, 0), dfs2(b.B, 0, 0);
            z[i] = b.A;
        }
    }
    for(int i=1; i<=n; ++i) son[f[i]].pb(i);
    for(int i=1; i<=n; ++i)
    {
        if(son[i].size())
        {
            for(auto j : son[i]) str[i].pb(dis[j]);
            sort(str[i].begin(), str[i].end());
            pre[i].pb((LL)str[i][0]);
            for(int j=1; j<str[i].size(); ++j) pre[i].pb(pre[i][j-1]+(LL)str[i][j]);
        }
    }
}
double cal(int a, int b)
{
    if(M.count(mp(a, b))) return M[mp(a,b)];
    double tmp = 0;
    for(auto i : str[a])
    {
        int pos = upper_bound(str[b].begin(), str[b].end(), max(z[a], z[b])-i-1)-str[b].begin();
        if(pos == str[b].size()) tmp += max(z[a], z[b])*1.0*son[b].size();
        else
            tmp += max(z[a], z[b])*1.0*pos + (i+1)*1.0*(son[b].size()-pos)+pre[b].back()-(pos==0?0:pre[b][pos-1]);
    }
    tmp /= (son[a].size()*1.0*son[b].size());
    M[mp(a,b)] = tmp;
    return tmp;
}
int main()
{
    int m, q;
    scanf("%d%d%d",&n,&m,&q);
    while(m--)
    {
        int a, b;
        scanf("%d%d",&a,&b);
        v[a].pb(b), v[b].pb(a);
    }
    init();
    while(q--)
    {
        int a, b;
        scanf("%d%d",&a,&b);
        if(f[a] == f[b]) {puts("-1"); continue;}
        if(son[f[a]].size() > son[f[b]].size()) swap(a, b);
        printf("%.10f\n",cal(f[a], f[b]));
    }
    return 0;
}