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<bits/stdc++.h>
using namespace std;

const int N=6100010;
int n,m,fa[N],ans[N],fv[N],u[N],v[N],x,y,hd[2010],cnt,Ans,tn[2],t[2][2010];
struct node{int to,next;}e[4010];
vector<int> V[2010];
bool bo[N];

void addedge(int x,int y)
{
	e[++cnt]=(node){y,hd[x]},hd[x]=cnt;
	e[++cnt]=(node){x,hd[y]},hd[y]=cnt;
}

void add(int x)
{
	V[u[x]].push_back(x);
	V[v[x]].push_back(x);
}

bool getpath(int x,int fa,int y,int o)
{
	if (x==y) return t[o][++tn[o]]=x,1;
	for (int i=hd[x]; i; i=e[i].next)
		if (e[i].to!=fa&&getpath(e[i].to,x,y,o)) return t[o][++tn[o]]=x,1;
	return 0;
}

void dfs(int x,int Fa)
{
	for (int i=hd[x]; i; i=e[i].next)
		if (e[i].to!=Fa) dfs(e[i].to,x);
	vector<int> vv; vv.clear();
	for (int i=0,sz=V[x].size(); i<sz; i++) 
		if (!bo[V[x][i]]) vv.push_back(V[x][i]),bo[V[x][i]]=1;
	Ans+=min((int)vv.size(),2);
	if (!vv.size()) return;
	if (vv.size()==1)
	{
		if ((u[vv[0]]^v[vv[0]]^x)==Fa) {ans[vv[0]]=0; return;}
		fa[vv[0]]=(++cnt),fv[vv[0]]=(u[vv[0]]==x);
		u[cnt]=(u[vv[0]]^v[vv[0]]^x),v[cnt]=Fa,add(cnt);
	} else
	{
		for (int j=2,sz=vv.size(); j<sz; j++)
		{
			if ((u[vv[j]]^v[vv[j]]^x)==Fa) {ans[vv[j]]=0; continue;}
			fa[vv[j]]=(++cnt),fv[vv[j]]=(u[vv[j]]==x);
			u[cnt]=(u[vv[j]]^v[vv[j]]^x),v[cnt]=Fa,add(cnt);
		}
		int p,p0=u[vv[0]]^v[vv[0]]^x,p1=u[vv[1]]^v[vv[1]]^x;
		tn[0]=0,getpath(x,0,p0,0),tn[1]=0,getpath(x,0,p1,1);
		while (tn[0]&&tn[1]&&t[0][tn[0]]==t[1][tn[1]]) p=t[0][tn[0]],tn[0]--,tn[1]--;
		if (p!=Fa) 
		{
			u[++cnt]=Fa,v[cnt]=p,add(cnt);
			u[++cnt]=Fa,v[cnt]=p,add(cnt);
		}
		if (p0!=p1)
		{
			fa[vv[0]]=fa[vv[1]]=(++cnt);
			fv[vv[0]]=(u[vv[0]]==x),fv[vv[1]]=(v[vv[1]]==x);
			u[cnt]=p0,v[cnt]=p1,add(cnt);
		} else ans[vv[0]]=0,ans[vv[1]]=(u[vv[0]]==u[vv[1]]);
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1; i<n; i++) scanf("%d%d",&x,&y),addedge(x,y);
	for (int i=1; i<=m; i++) scanf("%d%d",&u[i],&v[i]),add(i);
	Ans=0,cnt=m,dfs(1,0);
	for (int i=cnt; i; i--) if (fa[i]) ans[i]=ans[fa[i]]^fv[i];
	printf("%d\n",Ans);
	for (int i=1; i<=m; i++) if (ans[i]) printf("%d %d\n",v[i],u[i]); else printf("%d %d\n",u[i],v[i]);
	return 0;
}