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>
#define ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
const int N = 4010;
int dep[N],fa[N],tag[N],f[N],flg[N],n,m,u[N],v[N],ans,tot,T[N];
vector<int> e[N];
struct node{int u,v,num;};
vector<node> E;
void dfs(int x,int dad){
	fa[x]=dad;dep[x]=dep[dad]+1;
	for(int v:e[x]) if(v!=dad) dfs(v,x);
}
int lca(int x,int y){
	while(x!=y)
		if(dep[x]>dep[y]) x=fa[x];
		else y=fa[y];
	return x;
}
void solve(int x,int dad){
	for(int v:e[x]) if(v!=dad) solve(v,x);
	vector<node> V;
	for(auto&pr:E) if(pr.u==x||pr.v==x){
		if(pr.u==pr.v){pr.u=-1;pr.v=-1;continue;}
		V.push_back(pr);pr.u=pr.v=-1;
	}
	if(x==1) return;
	ans+=min(flg[x]+(int)V.size(),2);
	T[x]=min(flg[x]+(int)V.size(),2);
	for(int i=0;i+1<V.size();i+=2){
		if(V[i].v==x) tag[V[i].num]^=1,swap(V[i].u,V[i].v);
		if(V[i+1].v==x) tag[V[i+1].num]^=1,swap(V[i+1].u,V[i+1].v);
		node p={V[i].v,V[i+1].v,++tot};
		if(p.u==x) p.u=fa[x];
		if(p.v==x) p.v=fa[x];
		f[V[i].num]=p.num;f[V[i+1].num]=p.num;
		tag[V[i].num]^=1;
		E.push_back(p);
		int z=lca(p.u,p.v),z1=lca(V[i].u,V[i].v),z2=lca(V[i+1].u,V[i+1].v);
		if(z1==z2){
			if(z1==x) z1=fa[x];
			for(int y=x;y!=z1;y=fa[y]) flg[y]+=2;
			for(int y=z;y!=z1;y=fa[y]) flg[y]+=2;
		}
		else{
			if(dep[z1]<dep[z2]) z=z2;
			else z=z1;
			for(int y=x;y!=z;y=fa[y]) flg[y]+=2;
		}
	}
	if(V.size()&1){
		auto p=V.back();
		if(p.u==x) p.u=fa[x];
		if(p.v==x) p.v=fa[x];
		E.push_back(p);
	}
}
int main(){
	read(n);read(m);
	for(int i=1;i<n;i++){
		int x,y;read(x);read(y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs(1,0);
	for(int i=1;i<=m;i++){
		int x,y;read(x);read(y);
		u[i]=x;v[i]=y;
		E.push_back({x,y,i});
	}
	tot=m;
	solve(1,0);
	cout<<ans<<'\n';
	memset(flg,0,sizeof(flg));
	for(int i=1;i<=m;i++){
		int now=0,x=i;
		int z=lca(u[i],v[i]);
		for(int y=u[i];y!=z;y=fa[y]) flg[y]++;
		for(int y=v[i];y!=z;y=fa[y]) flg[y]++;
		while(x) now^=tag[x],x=f[x];
		if(now) cout<<v[i]<<' '<<u[i]<<'\n';
		else cout<<u[i]<<' '<<v[i]<<'\n';
	}
	int now=0;
	for(int i=1;i<=n;i++){
		now+=min(2,flg[i]);
	//	if(min(2,flg[i])!=T[i]) cout<<i<<' '<<flg[i]<<' '<<T[i]<<'\n';
	}
	return 0;
}
