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;

using ll=int64_t;
#define int ll

#define FOR(i,a,b) for(int i=int(a);i<int(b);i++)
#define REP(i,b) FOR(i,0,b)
#define MP make_pair
#define PB push_back
#define ALL(x) x.begin(),x.end()
#ifdef MAROON_LOCAL
#define cerr (cerr<<"-- line "<<__LINE__<<" -- ")
#else
class CerrDummy{}cerrDummy;
template<class T>
CerrDummy& operator<<(CerrDummy&cd,const T&){
	return cd;
}
using charTDummy=char;
using traitsDummy=char_traits<charTDummy>;
CerrDummy& operator<<(CerrDummy&cd,basic_ostream<charTDummy,traitsDummy>&(basic_ostream<charTDummy,traitsDummy>&)){
	return cd;
}
#define cerr cerrDummy
#endif
#define REACH cerr<<"reached"<<endl
#define DMP(x) cerr<<#x<<":"<<x<<endl
#define ZERO(x) memset(x,0,sizeof(x))
#define ONE(x) memset(x,-1,sizeof(x))

using pi=pair<int,int>;
using vi=vector<int>;
using ld=long double;

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"{";
	REP(i,(int)v.size()){
		if(i)os<<",";
		os<<v[i];
	}
	os<<"}";
	return os;
}

ll read(){
	ll i;
	scanf("%" SCNd64,&i);
	return i;
}

void printSpace(){
	printf(" ");
}

void printEoln(){
	printf("\n");
}

void print(ll x,int suc=1){
	printf("%" PRId64,x);
	if(suc==1)
		printEoln();
	if(suc==2)
		printSpace();
}

string readString(){
	static char buf[3341000];
	scanf("%s",buf);
	return string(buf);
}

char* readCharArray(){
	static char buf[3341000];
	static int bufUsed=0;
	char* ret=buf+bufUsed;
	scanf("%s",ret);
	bufUsed+=strlen(ret)+1;
	return ret;
}

template<class T,class U>
void chmax(T& a,U b){
	if(a<b)
		a=b;
}

template<class T,class U>
void chmin(T& a,U b){
	if(b<a)
		a=b;
}

template<class T>
T Sq(const T& t){
	return t*t;
}

#define CAPITAL
void Yes(bool ex=true){
	#ifdef CAPITAL
	cout<<"YES"<<endl;
	#else
	cout<<"Yes"<<endl;
	#endif
	if(ex)exit(0);
}
void No(bool ex=true){
	#ifdef CAPITAL
	cout<<"NO"<<endl;
	#else
	cout<<"No"<<endl;
	#endif
	if(ex)exit(0);
}

const ll infLL=LLONG_MAX/3;

#ifdef int
const int inf=infLL;
#else
const int inf=INT_MAX/2-100;
#endif

const int Nmax=2010;
const int Mmax=4010;
vector<pi> tr[Nmax];
int u[Mmax],v[Mmax],s[Mmax],t[Mmax],pr[Mmax][2],ans[Mmax],ord;
int otherEnd(int i,int x){
	return s[i]==x?t[i]:s[i];
}
bool merged[Mmax];
vi idx[Nmax];
void dfs(int cur,int p){
	for(auto e:tr[cur])if(e.first!=p)
		dfs(e.first,cur);
	int last=-1;
	for(auto i:idx[cur])if(!merged[i]){
		if(s[i]==t[i]){
			//do nothing
		}else{
			if(last==-1){
				last=i;
			}else{
				int j=ord++;
				u[j]=s[j]=otherEnd(last,cur);
				v[j]=t[j]=otherEnd(i,cur);
				merged[last]=true;
				merged[i]=true;
				pr[j][0]=last;
				pr[j][1]=i;
				idx[u[j]].PB(j);
				idx[v[j]].PB(j);
				last=-1;
			}
		}
	}
	if(last!=-1){
		assert(p!=-1);
		(s[last]==cur?s[last]:t[last])=p;
		idx[p].PB(last);
	}
}

int cnt[Nmax];
bool walk(int cur,int p,int tar){
	if(cur==tar)return true;
	for(auto e:tr[cur])if(e.first!=p){
		bool d=walk(e.first,cur,tar);
		if(d){
			cnt[e.second]++;
			return true;
		}
	}
	return false;
}

signed main(){
	int n=read(),m=read();
	REP(i,n-1){
		int a=read()-1,b=read()-1;
		tr[a].PB(pi(b,i));
		tr[b].PB(pi(a,i));
	}
	REP(i,m){
		u[i]=s[i]=read()-1;
		v[i]=t[i]=read()-1;
		idx[u[i]].PB(i);
		idx[v[i]].PB(i);
	}
	ord=m;
	dfs(0,-1);
	for(int i=ord-1;i>=m;i--){
		int a=pr[i][0];
		int b=pr[i][1];
		ans[a]=(ans[i]^(s[a]!=u[i]));
		ans[b]=(ans[i]^(t[b]!=v[i]));
	}
	REP(i,m)if(ans[i])
		swap(u[i],v[i]);
	REP(i,m){
		cerr<<u[i]<<" "<<v[i]<<endl;
		walk(u[i],-1,v[i]);
	}
	int fun=0;
	REP(i,n-1)
		fun+=min(cnt[i],int(2));
	print(fun);
	REP(i,m){
		print(u[i]+1,2);
		print(v[i]+1);
	}
}