Takahashi loves walking on a tree. The tree where Takahashi walks has N vertices numbered 1 through N. The i-th of the N-1 edges connects Vertex a_i and Vertex b_i.

Takahashi has scheduled M walks. The i-th walk is done as follows:

* The walk involves two vertices u_i and v_i that are fixed beforehand.
* Takahashi will walk from u_i to v_i or from v_i to u_i along the shortest path.



The happiness gained from the i-th walk is defined as follows:

* The happiness gained is the number of the edges traversed during the i-th walk that satisfies one of the following conditions:
* In the previous walks, the edge has never been traversed.
* In the previous walks, the edge has only been traversed in the direction opposite to the direction taken in the i-th walk.



Takahashi would like to decide the direction of each walk so that the total happiness gained from the M walks is maximized. Find the maximum total happiness that can be gained, and one specific way to choose the directions of the walks that maximizes the total happiness.

Constraints

* 1 ≤ N,M ≤ 2000
* 1 ≤ a_i , b_i ≤ N
* 1 ≤ u_i , v_i ≤ N
* a_i \neq b_i
* u_i \neq v_i
* The graph given as input 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}
u_1 v_1
:
u_M v_M


Output

Print the maximum total happiness T that can be gained, and one specific way to choose the directions of the walks that maximizes the total happiness, in the following format:


T
u^'_1 v^'_1
:
u^'_M v^'_M


where (u^'_i,v^'_i) is either (u_i,v_i) or (v_i,u_i), which means that the i-th walk is from vertex u^'_i to v^'_i.

Examples

Input

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


Output

6
2 3
3 4
4 2


Input

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


Output

6
2 4
3 5
5 1


Input

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


Output

9
2 4
6 3
5 6
4 5
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define mp make_pair
const int N=2050;
vector<int> E[N];
int n,deg[N],ans,par[N],dep[N],c[N];
void DFS(int u, int p)
{
	par[u]=p;
	dep[u]=dep[p]+1;
	for(int v:E[u]) if(v!=p) DFS(v,u);
}
void Set(int u, int v)
{
	if(dep[v]>dep[u]) swap(u,v);
	while(dep[u]>dep[v]) c[u]++,u=par[u];
	while(u!=v)
	{
		c[u]++;c[v]++;
		u=par[u];v=par[v];
	}
}
vector<pair<int,int> > Solve(vector<pair<int,int> > paths)
{
	//printf("Solve: ");
	//for(auto p:paths) printf("(%i %i) ",p.first,p.second);
	//printf("\n");
	if(paths.empty()) return paths;
	int u,v;
	for(u=1;u<=n;u++) if(deg[u]==1) break;
	if(u>n) return paths;
	for(int i=0;i<E[u].size();i++){ v=E[u][i];if(deg[v]) break;}
	//printf("u:%i v:%i\n",u,v);
	deg[u]--;deg[v]--;
	//printf(":D\n");
	vector<pair<int,int> > work,tmp;
	vector<int> id,pos(N,-1);
	for(int i=0;i<paths.size();i++)
	{
		if(paths[i].first==u || paths[i].second==u)
		{
			work.pb(paths[i]);
			id.pb(i);
		}
		else
		{
			pos[i]=tmp.size();
			tmp.pb(paths[i]);
		}
	}
	if(work.empty()) return Solve(paths);
	for(int i=0;i+1<work.size();i+=2)
	{
		if(work[i].second==u) swap(work[i].first,work[i].second);
		if(work[i+1].second==u) swap(work[i+1].first,work[i+1].second);
		if(work[i]==work[i+1])
		{
			if(paths[id[i]]==paths[id[i+1]]) swap(paths[id[i]].first,paths[id[i]].second);
			break;
		}
		pos[id[i]]=pos[id[i+1]]=tmp.size();
		tmp.pb(mp(work[i].second,work[i+1].second));
	}
	if(work.size()&1)
	{
		int i=work.size()-1;
		if(work[i].second==u) swap(work[i].first,work[i].second);
		if(work[i].second!=v)
		{
			pos[id[i]]=tmp.size();
			tmp.pb(mp(v,work[i].second));
		}
	}
	tmp=Solve(tmp);
	for(int i=0;i<paths.size();i++)
	{
		if(pos[i]!=-1 && tmp[pos[i]].first!=paths[i].first && tmp[pos[i]].second!=paths[i].second) swap(paths[i].first,paths[i].second);
	}
	return paths;
}
int main()
{
	int m;
	scanf("%i %i",&n,&m);
	int u,v;
	for(int i=1;i<n;i++) scanf("%i %i",&u,&v),E[u].pb(v),E[v].pb(u),deg[u]++,deg[v]++;
	vector<pair<int,int> > paths(m);
	DFS(1,0);
	for(int i=0;i<m;i++) scanf("%i %i",&paths[i].first,&paths[i].second),Set(paths[i].first,paths[i].second);
	for(int i=1;i<=n;i++) ans+=min(2,c[i]);
	paths=Solve(paths);
	printf("%i\n",ans);
	for(auto p:paths) printf("%i %i\n",p.first,p.second);
	return 0;
}