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=2005,S=41e5;
int u[S],v[S],faa[S],n,m,i,j,a,b,xb,id0,id1,ans,dfn[N],dcnt,dad[N],dep[N],q[N*2],qw;
bool cov[N];
vector<int>e[N];
void dfs(int x,int fa){dfn[++dcnt]=x;dad[x]=fa;dep[x]=dep[fa]+1;for(int y:e[x])if(y!=fa)dfs(y,x);}
inline int lca(int x,int y){return x==y?x:(dep[x]>dep[y]?lca(dad[x],y):lca(x,dad[y]));}
inline void covchain(int b,int a){for(;b!=a;b=dad[b])cov[b]=1;}
inline void work(int x){
	int ii,i;
	for(ii=1;ii<=qw;++ii)if(i=q[ii],u[i]!=x)swap(u[i],v[i]);
	id0=id1=0;if(!cov[x])for(ii=1;ii<=qw;++ii)if(i=q[ii],!faa[i]){if(u[i]!=x)swap(u[i],v[i]);if(u[i]==x)(id0?id1:id0)=i;}
	if(id1){
		if(::i==2){if(v[id0]==v[id1])swap(u[id0],v[id0]);cov[x]=1;return;}
		faa[id0]=faa[id1]=++xb;u[xb]=u[id0]^v[id0]^x;v[xb]=u[id1]^v[id1]^x;
		a=lca(x,u[xb]);b=lca(x,v[xb]);
		if(a==b)covchain(lca(u[xb],v[xb]),a),covchain(x,a);
			else covchain(x,dep[a]>dep[b]?a:b);
		if(u[xb]==v[xb]){faa[xb]=-1;if(u[id0]==u[id1])swap(u[id0],v[id0]);}
			else q[++qw]=xb;
	}else if(id0)++ans;
	for(ii=1;ii<=qw;++ii)if(i=q[ii],!faa[i] && u[i]==x){
		if(v[i]!=dad[x])q[++qw]=faa[i]=++xb,u[xb]=dad[x],v[xb]=v[i];
			else faa[i]=-1;
	}
	int lim=qw;qw=0;
	for(ii=1;ii<=lim;++ii)if(!faa[q[ii]])q[++qw]=q[ii];
}
int main(){
	scanf("%d%d",&n,&m);xb=qw=m;for(i=1;i<=qw;++i)q[i]=i;
	for(i=1;i<n;++i)scanf("%d%d",&a,&b),e[a].push_back(b),e[b].push_back(a);
	for(i=1;i<=m;++i)scanf("%d%d",u+i,v+i);
	dfs(1,0);
	for(i=n;i>1;--i)work(dfn[i]);
	for(i=2;i<=n;++i)ans+=cov[dfn[i]]*2;
	for(i=xb;i;--i)if(faa[i]>0 && (u[i]==v[faa[i]] || v[i]==u[faa[i]]))swap(u[i],v[i]);
	printf("%d\n",ans);
	for(i=1;i<=m;++i)printf("%d %d\n",u[i],v[i]);
	return 0;
}