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<cstdio>
using namespace std;
#define N 4050
int n,m,st[N],fg,head[N],cnt,vl[N],ct,s,t,as,v[N][2],ct2,is[N],nt[N],v2[N][2];
struct edge{int t,next,id;}ed[N*2];
struct sth{int f,t,v;};bool operator <(sth a,sth b){return a.v<b.v;}
void adde(int f,int t,int id){ed[++cnt]=(edge){t,head[f],id};head[f]=cnt;ed[++cnt]=(edge){f,head[t],id};head[t]=cnt;}
void adde2(int f,int t,int id){ed[++cnt]=(edge){t,head[f],id*2};head[f]=cnt;ed[++cnt]=(edge){f,head[t],id*2+1};head[t]=cnt;}
void dfs(int u,int fa)
{
	if(!fg)st[++ct]=u;
	if(u==t)fg=1;
	for(int i=head[u];i;i=ed[i].next)if(ed[i].t!=fa)dfs(ed[i].t,u);
	if(!fg)ct--;
}
void dfs2(int u,int fa)
{
	for(int i=head[u];i;i=ed[i].next)if(ed[i].t!=fa)dfs2(ed[i].t,u);
	int ls=0;
	for(int i=1;i<=ct2;i++)
	if(v[i][0]==u||v[i][1]==u)
	{
		if(!ls)ls=i;
		else
		{
			int l=v[ls][0],r=v[i][1];
			if(v[ls][0]==u)l=v[ls][1],is[ls]=1;
			if(v[i][1]==u)r=v[i][0],is[i]=1;
			nt[ls]=nt[i]=++ct2;
			v[ct2][0]=l;v[ct2][1]=r;
			v[ls][0]=v[ls][1]=v[i][0]=v[i][1]=0;ls=0;
			if(v[ct2][0]==v[ct2][1])v[ct2][0]=v[ct2][1]=0;
		}
	}
	if(ls)
	{
		if(v[ls][0]==u)v[ls][0]=fa;
		else v[ls][1]=fa;
		if(v[ls][0]==v[ls][1])v[ls][0]=v[ls][1]=0;
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<n;i++)scanf("%d%d",&s,&t),adde(s,t,i);
	for(int r=1;r<=m;r++)
	{
		scanf("%d%d",&s,&t);ct=fg=0;dfs(s,0);v[r][0]=s,v[r][1]=t;v2[r][0]=s;v2[r][1]=t;
		for(int i=1;i<ct;i++)for(int j=head[st[i]];j;j=ed[j].next)if(ed[j].t==st[i+1]&&vl[ed[j].id]<2)vl[ed[j].id]++,as++;
	}
	printf("%d\n",as);
	ct2=m;dfs2(1,0);
	for(int i=ct2;i>=1;i--)
	is[i]^=is[nt[i]];
	for(int i=1;i<=m;i++)if(is[i])printf("%d %d\n",v2[i][1],v2[i][0]);
	else printf("%d %d\n",v2[i][0],v2[i][1]);
}