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=2010;
int cnt,n,m,ans,f[N],fu[N*2],d[N],del[N*2],vis[N],tp[N*2],a[N*2],b[N*2],tf[N*2];
list<int>L[N*2],li[N*2];
vector<int>g[N];
queue<int>q;
int find(int x){return f[x]==x?x:f[x]=find(f[x]);}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i=1;i<n;++i){
		int x,y;cin>>x>>y;
		g[x].push_back(y);g[y].push_back(x);
		++d[x];++d[y];
	}
	for(int i=1;i<=m;++i){
		cin>>a[i]>>b[i];
		li[a[i]].push_back(i);li[b[i]].push_back(i);
		L[a[i]].push_back(i);L[b[i]].push_back(i);
	}
	cnt=m;
	for(int i=1;i<=n;i++)if(d[i]<=1)q.push(i);
	for(int i=1;i<=n;i++)f[i]=i;
	while(!q.empty()){
		int x=q.front();q.pop();vis[x]=1;
		int u=-1,v=-1;
		for(;!li[x].empty();){
			int t=li[x].back();li[x].pop_back();
			if(del[t])continue;
			if(find(a[t])==find(b[t])){del[t]=1;tp[t]=0;continue;}
			if(u==-1)u=t;
			else{v=t;break;}
		}
		if(v==-1){if(u!=-1)li[x].push_back(u);}
		else{
			++cnt;a[cnt]=find(a[u])+find(b[u])-x;b[cnt]=find(a[v])+find(b[v])-x;
			li[a[cnt]].push_back(cnt);li[b[cnt]].push_back(cnt);
			fu[u]=fu[v]=cnt;del[u]=del[v]=1;
			if(find(b[u])==x)tf[u]=0;else tf[u]=1;
			if(find(a[v])==x)tf[v]=0;else tf[v]=1;
		}
		u=v=-1;
		for(;!L[x].empty();){
			int t=L[x].back();L[x].pop_back();
			if(find(a[t])==find(b[t]))continue;
			if(u==-1)u=t;
			else{v=t;break;}
		}
		if(u!=-1)++ans,L[x].push_back(u);if(v!=-1)++ans,L[x].push_back(v);
		for(int i:g[x])
			if(!vis[i]){
				f[x]=i;li[i].splice(li[i].end(),li[x]);L[i].splice(L[i].end(),L[x]);
				--d[i];if(d[i]==1)q.push(i);
			}
	}
	cout<<ans<<endl;
	for(int i=cnt;i;i--)if(fu[i])tp[i]=tp[fu[i]]^tf[i];
	for(int i=1;i<=m;++i){
		if(tp[i])swap(a[i],b[i]);
		cout<<a[i]<<' '<<b[i]<<'\n';
	}
	return 0;
}