Problem: 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

Solution (in CPP):

Solution:
using namespace std;
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#define N 2010
int n,m;
struct EDGE{
	int to;
	EDGE *las;
};
struct Graph{
	EDGE *last[N*2];
	void link(int u,int v){
		EDGE *nw=new EDGE;
		*nw={v,last[u]};
		last[u]=nw;
	}
} G,F,L;
int d[N];
struct edge{
	int u,v;
} og[N],p[N*2];
int np;
#define Lto(ei) (p[ei->to].u^p[ei->to].v^(x))
void Flk(int u,int v,int ty){
	F.link(u,v*ty);
	F.link(v,u*ty);
}
int anss;
int ans[N*2];
set<pair<int,int> > s;
int fa[N],dep[N];
void dfs(int x){
	dep[x]=dep[fa[x]]+1;
	for (EDGE *ei=G.last[x];ei;ei=ei->las)
		if (ei->to!=fa[x]){
			fa[ei->to]=x;
			dfs(ei->to);
		}
}
int lca(int a,int b){
	if (dep[a]<dep[b]) swap(a,b);
	for (int k=dep[a]-dep[b];k;--k)
		a=fa[a];
	for (;a!=b;a=fa[a],b=fa[b]);
	return a;
}
void topu(){
	static int q[N];
	static bool vis[N];
	int head=1,tail=0;
	for (int i=1;i<=n;++i)
		if (d[i]==1)
			q[++tail]=i;
	while (head<=tail){
		int x=q[head++],y=0;
		vis[x]=1;
		for (EDGE *ei=G.last[x];ei;ei=ei->las)
			if (!vis[ei->to])
				y=ei->to;
		if (!y) continue;
		if (--d[y]==1)
			q[++tail]=y;
		int cnt=0;
		for (EDGE *ei=L.last[x];ei;ei=ei->las)
			if (!vis[Lto(ei)])
				cnt++;
		if (cnt==1 || s.find(make_pair(x,y))!=s.end()){
			anss++;
			for (EDGE *ei=L.last[x];ei;ei=ei->las)
				if (!vis[Lto(ei)]){
					(p[ei->to].u==x?p[ei->to].u:p[ei->to].v)=y;
					L.link(y,ei->to);
				}
		}
		else if (cnt>1){
			EDGE *e1=NULL,*e2=NULL,*ei;
			for (ei=L.last[x];ei && !e2;ei=ei->las)
				if (!vis[Lto(ei)]){
					if (!e1) e1=ei;
					else e2=ei;
				}
			for (;ei;ei=ei->las)
				if (!vis[Lto(ei)]){
					(p[ei->to].u==x?p[ei->to].u:p[ei->to].v)=y;
					L.link(y,ei->to);
				}
			fa[x]=0,dfs(x);
			int a=Lto(e1),b=Lto(e2),z=lca(a,b);
			p[++np]={a,b};
			L.link(a,np),L.link(b,np);
			Flk(np,e1->to,p[e1->to].u==a?1:-1);
			Flk(e1->to,e2->to,(int(p[e1->to].u==x) ^ int(p[e2->to].u==x))?1:-1);
			for (;z!=x;z=fa[z]){
				if (s.find(make_pair(z,fa[z]))!=s.end())
					continue;
				s.insert(make_pair(z,fa[z]));
				s.insert(make_pair(fa[z],z));
				anss++;
			}
			anss++;
		}
	}
}
void color(int x){
	for (EDGE *ei=F.last[x];ei;ei=ei->las){
		int y=abs(ei->to);
		if (!ans[y]){
			ans[y]=(ei->to>0?ans[x]:-ans[x]);
			color(y);
		}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<n;++i){
		int u,v;
		scanf("%d%d",&u,&v);
		G.link(u,v),G.link(v,u);
		d[u]++,d[v]++;
	}
	for (int i=1;i<=m;++i){
		int x,y;
		scanf("%d%d",&x,&y);
		p[++np]={x,y};
		og[np]=p[np];
		L.link(x,np),L.link(y,np);
	}
	topu();
	printf("%d\n",anss);
	for (int i=1;i<=m;++i){
		if (ans[i]==0)
			ans[i]=1,color(i);
		if (ans[i]==1)
			printf("%d %d\n",og[i].u,og[i].v);
		else
			printf("%d %d\n",og[i].v,og[i].u);
	}
	return 0;
}