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>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
std::mt19937 rnd(time(NULL));
#define rand rnd
template<class T>void cxk(T&a,T b){a=a>b?a:b;}
template<class T>void cnk(T&a,T b){a=a<b?a:b;}
#ifdef mod
int pow(int x,int y){
	int ret=1;
	while(y){
		if(y&1)ret=1ll*ret*x%mod;
		x=1ll*x*x%mod;y>>=1;
	}
	return ret;
}
template<class Ta,class Tb>void inc(Ta&a,Tb b){a=a+b>=mod?a+b-mod:a+b;}
template<class Ta,class Tb>void dec(Ta&a,Tb b){a=a>=b?a-b:a+mod-b;}
#endif
int fir[2010],dis[4010],nxt[4010],id;
void link(int a,int b){nxt[++id]=fir[a],fir[a]=id,dis[id]=b;}
int lp[4010],rp[4010],lps[4010],rps[4010],del[4010],M,pa[4010],pb[4010],dir[4010];
struct pr{int fi,se,th;};
std::vector<pr>v[2010];
int n,m,FA[2010],ans;
int hd(int x){return FA[x]==x?x:FA[x]=hd(FA[x]);}
std::bitset<2001>dfs(int x,int fa=-1){
	std::bitset<2001>ret;
	for(int i=1;i<=m;++i)if(pa[i]==x||pb[i]==x)ret[i]=1;
	for(int i=fir[x];i;i=nxt[i])if(dis[i]!=fa)ret^=dfs(dis[i],x),FA[dis[i]]=x;
	if(x==1)return ret;
	for(int i=1;i<=M;++i)if(hd(pa[i])==hd(pb[i]))del[i]=1;
	ans+=std::min(2,(int)ret.count());
	std::vector<pr>_;
	for(pr i:v[x])if(!del[i.fi])_.push_back(i);
	v[x]=_;
	if(v[x].size()>=2){
		pr a=v[x][0],b=v[x][1];
		v[x].erase(v[x].begin(),v[x].begin()+2);
		a.se=hd(a.se),b.se=hd(b.se);
		++M;pa[M]=a.se,pb[M]=b.se;
		lp[M]=a.fi,rp[M]=b.fi;
		lps[M]=!a.th,rps[M]=b.th;
		del[a.fi]=del[b.fi]=1;
		v[a.se].push_back({M,b.se,1});
		v[b.se].push_back({M,a.se,0});
	}
	if(~fa)for(pr i:v[x])v[fa].push_back(i);
	return ret;
}
int main(){
#ifdef LOCAL
	freopen("in.in","r",stdin);
	//freopen("out.out","w",stdout);
#endif
	n=gi(),m=gi();int a,b;M=m;
	for(int i=1;i<=n;++i)FA[i]=i;
	for(int i=1;i<n;++i)a=gi(),b=gi(),link(a,b),link(b,a);
	for(int i=1;i<=m;++i)pa[i]=gi(),pb[i]=gi(),v[pa[i]].push_back({i,pb[i],1}),v[pb[i]].push_back({i,pa[i],0});
	dfs(1);
	for(int i=M;i;--i){
		if(!dir[i])dir[i]=1;
		dir[lp[i]]=lps[i]?dir[i]:3-dir[i];
		dir[rp[i]]=rps[i]?dir[i]:3-dir[i];
	}
	printf("%d\n",ans);
	for(int i=1;i<=m;++i)
		if(dir[i]==1)printf("%d %d\n",pa[i],pb[i]);
		else assert(dir[i]==2),printf("%d %d\n",pb[i],pa[i]);
	return 0;
}