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 <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 4000010
pair <int,int> edge[N],edge1[N],tree[N]; vector <int> tmp1,tmp2; bool need[N];
int ans=0,many[N],in[N],n,m,tot,level[N],fa[N],head[N],to[N<<1],nxt[N<<1],ff[N],idx;
void add(int a,int b) {nxt[++idx]=head[a],to[idx]=b,head[a]=idx;}
void dfs(int p,int from)
{
	fa[p]=from,level[p]=level[from]+1;
	for(int i=head[p];i;i=nxt[i]) if(to[i]!=from) dfs(to[i],p);
}
void calc(int x,int y)
{
	if(level[x]>level[y]) swap(x,y);
	while(level[y]>level[x]) many[y]++,y=fa[y];
	while(x!=y) many[x]++,many[y]++,x=fa[x],y=fa[y];
}
void solve(vector <int> tr,vector <int> ed)
{
	//for(int i=0;i<(int)tr.size();i++) printf("%d ",tr[i]); printf("\n");
	//for(int i=0;i<(int)ed.size();i++) printf("%d ",ed[i]); printf("\n");
	if(!tr.size()) return; if(ed.size()<=1) {for(int i=0;i<(int)ed.size();i++) need[ed[i]]=false;return;}
	for(int i=1;i<=n;i++) in[i]=0;
	for(int i=0;i<(int)tr.size();i++) in[tree[tr[i]].first]++,in[tree[tr[i]].second]++;
	int fr=0,to=0; vector <int> new_edge,new_tree;
	for(int i=1;i<=n;i++) if(in[i]==1) {fr=i;break;}
	for(int i=0;i<(int)tr.size();i++)
		if(tree[tr[i]].first==fr||tree[tr[i]].second==fr) to=tree[tr[i]].first^tree[tr[i]].second^fr;
		else new_tree.push_back(tr[i]);
	vector <int> vis(2*m+1); int j=-1;
	for(int i=0;i<(int)ed.size();i++)
		if(edge[ed[i]].first==fr||edge[ed[i]].second==fr)
		{
			if(j!=-1)
			{
				int a=edge[ed[j]].first^edge[ed[j]].second^fr,b=edge[ed[i]].first^edge[ed[i]].second^fr;
				if(a!=b)
				{
					ff[ed[i]]=ff[ed[j]]=++tot,new_edge.push_back(tot),
					edge[tot]=make_pair(a,b),need[ed[j]]=edge[ed[j]].first!=a,need[ed[i]]=edge[ed[i]].second!=b;
					vis[ed[i]]=vis[ed[j]]=true;
				} else need[ed[i]]=true,need[ed[j]]=edge[ed[j]].first!=edge[ed[i]].first;
				j=-1;
			} else j=i;
		}
		else new_edge.push_back(ed[i]);
	if(j!=-1)
	{
		if(edge[ed[j]].first==fr) edge[ed[j]].first=to;
		if(edge[ed[j]].second==fr) edge[ed[j]].second=to;
		if(edge[ed[j]].first!=edge[ed[j]].second) new_edge.push_back(ed[j]);
	}
	solve(new_tree,new_edge);
	for(int i=0;i<(int)ed.size();i++) if(vis[ed[i]]) need[ed[i]]^=need[ff[ed[i]]];
}
int main()
{
	scanf("%d%d",&n,&m),tot=m;
	for(int i=1,a,b;i<n;i++) scanf("%d%d",&a,&b),add(a,b),add(b,a),tree[i]=make_pair(a,b),tmp1.push_back(i);
	for(int i=1,a,b;i<=m;i++) scanf("%d%d",&a,&b),edge1[i]=edge[i]=make_pair(a,b),tmp2.push_back(i);
	dfs(1,0);
	for(int i=1;i<=m;i++) calc(edge[i].first,edge[i].second);
	for(int i=1;i<=n;i++) ans+=min(many[i],2);
	printf("%d\n",ans),solve(tmp1,tmp2);
	for(int i=1;i<=m;i++)
		if(!need[i]) printf("%d %d\n",edge1[i].first,edge1[i].second);
		else printf("%d %d\n",edge1[i].second,edge1[i].first);
}