Problem: You are given a tree T with N vertices and an undirected graph G with N vertices and M edges. The vertices of each graph are numbered 1 to N. The i-th of the N-1 edges in T connects Vertex a_i and Vertex b_i, and the j-th of the M edges in G connects Vertex c_j and Vertex d_j.

Consider adding edges to G by repeatedly performing the following operation:

* Choose three integers a, b and c such that G has an edge connecting Vertex a and b and an edge connecting Vertex b and c but not an edge connecting Vertex a and c. If there is a simple path in T that contains all three of Vertex a, b and c in some order, add an edge in G connecting Vertex a and c.



Print the number of edges in G when no more edge can be added. It can be shown that this number does not depend on the choices made in the operation.

Constraints

* 2 \leq N \leq 2000
* 1 \leq M \leq 2000
* 1 \leq a_i, b_i \leq N
* a_i \neq b_i
* 1 \leq c_j, d_j \leq N
* c_j \neq d_j
* G does not contain multiple edges.
* T is a tree.

Input

Input is given from Standard Input in the following format:


N M
a_1 b_1
:
a_{N-1} b_{N-1}
c_1 d_1
:
c_M d_M


Output

Print the final number of edges in G.

Examples

Input

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


Output

6


Input

7 5
1 5
1 4
1 7
1 2
2 6
6 3
2 5
1 3
1 6
4 6
4 7


Output

11


Input

13 11
6 13
1 2
5 1
8 4
9 7
12 2
10 11
1 9
13 7
13 11
8 10
3 8
4 13
8 12
4 7
2 3
5 11
1 4
2 11
8 10
3 5
6 9
4 10


Output

27

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> ii;

int n, m; 
vector<int> tree[2005];
int dist[2005][2005];
int parent[2005][2005]; ///parent[root of tree][node]
bool collinear(int a, int b, int c){
	int X = dist[a][b];
	int Y = dist[b][c];
	int Z = dist[a][c];
	if(max(max(X,Y),Z) * 2 == X+Y+Z) return true;
	else return false;
}
void dfs1(int u, int p, int root){
	parent[root][u] = p;
	for(int v : tree[u]){
		if(v == p) continue;
		dist[root][v] = dist[root][u] + 1;
		dfs1(v, u, root);
	}
}

bool G[2005][2005]; ///shortened graph
vector<int> adj[2005]; ///shortened graph but adjList

///let c be trigger[a][b]. if a-b is added, a-c shortens to b-c.
///i.e. if the tree is rooted at a, b is in the subtree of c
int trigger[2005][2005]; 
void add(int a, int b){
	
	if(trigger[a][b] == b || trigger[b][a] == a) return; ///edge alrd exists
	else if(trigger[a][b]) add(trigger[a][b], b); ///shorten a-b to b-c
	else if(trigger[b][a]) add(trigger[b][a], a); ///shorten a-b to a-c
	else{
		///this is to settle the case where a-b shortens some other edge
		trigger[a][b] = b;
		trigger[b][a] = a;
		G[a][b] = 1; G[b][a] = 1; ///add the edge
		
		vector<ii> toadd;
		queue<int> bfs;
		
		bfs.push(b);
		while(!bfs.empty()){ ///explore the subtree of b (where a is root)
			int u = bfs.front(); bfs.pop();
			for(int v : tree[u]){
				if(v == parent[a][u]) continue; ///subtree only
				
				if(trigger[a][v] == 0){
					trigger[a][v] = b;
					bfs.push(v);
				}
				else{
					G[a][v] = 0; G[v][a] = 0; ///break the old edge
					toadd.push_back(ii(b, v)); 
				}
			}
		}
		
		bfs.push(a);
		while(!bfs.empty()){ ///explore the subtree of a (where b is root)
			int u = bfs.front(); bfs.pop();
			for(int v : tree[u]){
				if(v == parent[b][u]) continue; ///subtree only
				
				if(trigger[b][v] == 0){
					trigger[b][v] = a;
					bfs.push(v);
				}
				else{
					G[b][v] = 0; G[v][b] = 0; ///break the old edge
					toadd.push_back(ii(a, v)); 
				}
			}
		}
		
		for(ii e : toadd) add(e.first, e.second);
	}
}


int ans = 0;
void dfs2(int u, int p){
	for(int v : adj[u]){
		if(v == p) continue;
		if(p != -1 && !collinear(u,v,p)) continue;
		
		ans++;
		dfs2(v,u);
	}
}

int main(){
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n >> m;
	
	for(int i = 1;i < n;i++){
		int a, b; cin >> a >> b;
		tree[a].push_back(b);
		tree[b].push_back(a);
	}
	for(int i = 1;i <= n;i++) dfs1(i, -1, i);
	
	for(int i = 1;i <= m;i++){
		int a, b; cin >> a >> b;
		add(a,b); ///add the edges one by one
	}
	
	for(int u = 1;u <= n;u++){
		for(int v = 1;v <= n;v++){
			if(G[u][v]) adj[u].push_back(v); ///convert adjmat to adjlist
		}
	}
	
	for(int i = 1;i <= n;i++) dfs2(i,-1);
	cout << ans/2;
}