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<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#define LL long long
#define LD long double
using namespace std;
const int NN=4000 +117;
const int MM=2000 +117;
int read(){
	int fl=1,x;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if(c=='-'){fl=-1;c=getchar();}
	for(x=0;c>='0'&&c<='9';c=getchar())
		x=(x<<3)+(x<<1)+c-'0';
	return x*fl;
}
void open(){
	freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
}
void close(){
	fclose(stdin);
	fclose(stdout);
}


int m,n;
vector<int> p[NN]={};
int fa[NN]={};
int dep[NN]={};
vector<int> t[NN]={};
int te[NN][2]={};
int e[NN][2]={};
int trans[NN]={};
int ecnt=0;
int val[NN]={};
void dfs(int x,int f){
	dep[x]=dep[f]+1;
	fa[x]=f;
	for(int i=0;i<p[x].size();++i){
		int cur=p[x][i];
		if(cur!=f){
			dfs(cur,x);
		}
	}
}
void walk(int x,int y,int id){
	while(x!=y){
		if(dep[x]<dep[y])swap(x,y);
		t[x].push_back(id);
		x=fa[x];
	}
}
vector<int> tp;
void calc(int x){
	for(int i=0;i<p[x].size();++i){
		int cur=p[x][i];
		if(cur!=fa[x])calc(cur);
	}
	tp.clear();
	for(int i=0;i<t[x].size();++i){
		if(!trans[t[x][i]]&&e[t[x][i]][0]!=e[t[x][i]][1])tp.push_back(t[x][i]);
	}
	if(tp.size()>=2){
		int a=*tp.rbegin();
		int b=*(tp.rbegin()+1);
		++ecnt;
		if(e[a][0]==x){
			val[a]=1;
			e[ecnt][0]=e[a][1];
		}
		else e[ecnt][0]=e[a][0];
		if(e[b][1]==x){
			val[b]=1;
			e[ecnt][1]=e[b][0];
		}
		else e[ecnt][1]=e[b][1];
		walk(e[ecnt][0],e[ecnt][1],ecnt);
		trans[a]=ecnt;
		trans[b]=ecnt;
		tp.erase(tp.end()-1);
		tp.erase(tp.end()-1);
	}
	for(int i=0;i<tp.size();++i){
		int k=tp[i];
		if(e[k][0]==x)e[k][0]=fa[x];
		else e[k][1]=fa[x];
	}
}
int main(){
	//open();
	n=read();
	m=read();
	for(int i=1;i<n;++i){
		int x=read();
		int y=read();
		p[x].push_back(y);
		p[y].push_back(x);
	}
	dfs(1,0);
	for(int i=1;i<=m;++i){
		te[i][0]=e[i][0]=read();
		te[i][1]=e[i][1]=read();
		walk(e[i][1],e[i][0],i);
	}
	ecnt=m;
	int ans=0;
	for(int i=2;i<=n;++i){
		ans+=min((int)t[i].size(),2);
	}
	calc(1);
	for(int i=ecnt;i;--i){
		val[i]=val[i]^val[trans[i]];
	}
	printf("%d\n",ans);
	for(int i=1;i<=m;++i){
		if(val[i])printf("%d %d\n",te[i][1],te[i][0]);
		else printf("%d %d\n",te[i][0],te[i][1]);
	}
	close();
	return 0;
}