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<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
int h[2010],nex[4010],to[4010],dep[2010],fa[2010],fav[2010],M;
void add(int a,int b){
	M++;
	to[M]=b;
	nex[M]=h[a];
	h[a]=M;
}
void dfs(int x){
	dep[x]=dep[fa[x]]+1;
	for(int i=h[x];i;i=nex[i]){
		if(to[i]!=fa[x]){
			fa[to[i]]=x;
			dfs(to[i]);
		}
	}
}
void col(int x,int y){
	if(dep[x]<dep[y])swap(x,y);
	while(dep[x]>dep[y]){
		fav[x]++;
		x=fa[x];
	}
	while(x!=y){
		fav[x]++;
		fav[y]++;
		x=fa[x];
		y=fa[y];
	}
}
struct rt{
	int x,y;
	rt(int a=0,int b=0){x=a;y=b;}
};
bool operator==(rt a,rt b){return a.x==b.x&&a.y==b.y;}
int d[2010],n;
vector<rt>solve(vector<rt>v){
	if(v.size()==0)return v;
	int x,u,i;
	for(x=1;x<=n;x++){
		if(d[x]==1)break;
	}
	if(x>n)return v;
	for(i=h[x];i;i=nex[i]){
		if(d[to[i]]){
			u=to[i];
			break;
		}
	}
	d[x]--;
	d[u]--;
	vector<rt>lx,nv;
	vector<int>id;
	int pos[2010];
	memset(pos,-1,sizeof(pos));
	for(i=0;i<(int)v.size();i++){
		if(v[i].x==x||v[i].y==x){
			lx.push_back(v[i]);
			id.push_back(i);
		}else{
			nv.push_back(v[i]);
			pos[i]=nv.size()-1;
		}
	}
	if(lx.size()==0)return solve(v);
	for(i=0;i+1<(int)lx.size();i+=2){
		if(lx[i].y==x)swap(lx[i].x,lx[i].y);
		if(lx[i+1].y==x)swap(lx[i+1].x,lx[i+1].y);
		if(lx[i]==lx[i+1]){
			if(v[id[i]]==v[id[i+1]])swap(v[id[i]].x,v[id[i]].y);
			continue;
		}
		nv.push_back(rt(lx[i].y,lx[i+1].y));
		pos[id[i]]=pos[id[i+1]]=nv.size()-1;
	}
	if(lx.size()&1){
		if(lx[lx.size()-1].y==x)swap(lx[lx.size()-1].x,lx[lx.size()-1].y);
		if(lx[lx.size()-1].y!=u){
			nv.push_back(rt(u,lx[lx.size()-1].y));
			pos[id[lx.size()-1]]=nv.size()-1;
		}
	}
	nv=solve(nv);
	for(i=0;i<(int)v.size();i++){
		if(pos[i]!=-1&&!(v[i].x==nv[pos[i]].x||v[i].y==nv[pos[i]].y))swap(v[i].x,v[i].y);
	}
	return v;
}
vector<rt>v;
int main(){
	int m,i,x,y,ans;
	scanf("%d%d",&n,&m);
	for(i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		add(x,y);
		add(y,x);
		d[x]++;
		d[y]++;
	}
	dfs(1);
	for(i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		v.push_back(rt(x,y));
		col(x,y);
	}
	v=solve(v);
	ans=0;
	for(i=2;i<=n;i++)ans+=min(2,fav[i]);
	printf("%d\n",ans);
	for(rt t:v)printf("%d %d\n",t.x,t.y);
}