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:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m;
int s[2010],t[2010];
struct Edge{
	int v,nxt;
}e[4010];int tot=1;
int first[2010],deg[2010];
void build(int u,int v)
{
	e[++tot]=(Edge){v,first[u]};
	first[u]=tot;deg[u]++;
	return ;
}
inline int eid(int x){return x/2;}
int c[2010];
int vis[2010][2010];
bool dfs(int u,int lst,int tar,int id)
{
	if(u==tar)return true;
	for(int i=first[u];i;i=e[i].nxt)
		if(e[i].v!=lst){
			bool flag=dfs(e[i].v,u,tar,id);
			if(flag){
				vis[id][eid(i)]=i&1;
				c[eid(i)]++;
				return true;
			}
		}
	return false;
}
void init()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d %d",&u,&v);
		build(u,v);
		build(v,u);
	}
	memset(vis,-1,sizeof(vis));
	for(int i=1;i<=m;i++)
	{
		scanf("%d %d",&s[i],&t[i]);
		dfs(s[i],0,t[i],i);
	}
	int ans=0;
	for(int i=1;i<n;i++)
		ans+=min(c[i],2);
	printf("%d\n",ans);
	return ;
}
int fa[2010],val[2010];
int find(int x)
{
	if(fa[x]==x)return x;
	int f=find(fa[x]);
	val[x]^=val[fa[x]];
	return fa[x]=f;
}
queue<int>q;
bool book[2010];
void work(int u)
{
	book[u]=true;
	for(int i=first[u];i;i=e[i].nxt)
	{
		deg[e[i].v]--;
		if(deg[e[i].v]==1)q.push(e[i].v);
		if(c[eid(i)]>=2)
		{
			int t=eid(i);
			int a=0,b=0;
			for(int o=1;o<=m;o++)
				if(fa[o]==o&&vis[o][t]!=-1)
				{
					if(!a)a=o;
					else{
						b=o;
						break;
					}
				}
			int tmp=vis[a][t]^vis[b][t]^1;
			fa[b]=a;val[b]=tmp;
			for(int i=1;i<n;i++)
				if(vis[b][i]!=-1){
					if(vis[a][i]!=-1)vis[a][i]=-1,c[i]=0;
					else vis[a][i]=vis[b][i]^tmp;
				}
		}
	}
	return ;
}
void solve()
{
	for(int i=1;i<=m;i++)
		fa[i]=i,val[i]=0;
	for(int i=1;i<=n;i++)
		if(deg[i]==1)q.push(i);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		work(u);
	}
	for(int i=1;i<=m;i++)
	{
		find(i);
		if(val[i])swap(s[i],t[i]);
		printf("%d %d\n",s[i],t[i]);
	}
	return ;
}
int main()
{
	init();
	solve();
	return 0;
}