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
#include<bits/stdc++.h>
using namespace std;
const int N=4004;
struct node{int to,next;}num[N];
int cnt=1,n,m,u,v,ans,Cnt[N],st[N],ed[N],head[N],vis[N],sure[N];
vector<int> vec[N],pass[N]; 
void add(int x,int y)
{num[++cnt].to=y;num[cnt].next=head[x];head[x]=cnt;}
int dfs(int x,int fa,int aim,int id)
{
	if (x==aim) return 1;
	for (int i=head[x];i;i=num[i].next)
	  if (num[i].to!=fa)
	  {
	  	  Cnt[i/2]++;pass[i/2].push_back(id);
	  	  vec[id].push_back(i);
	  	  if (dfs(num[i].to,x,aim,id)==1) return 1;	
		  Cnt[i/2]--;pass[i/2].pop_back();vec[id].pop_back();
	  }
	return 0;
}
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<n;i++) scanf("%d%d",&u,&v),add(u,v),add(v,u);
	for (int i=1;i<=m;i++)
	{
		scanf("%d%d",&st[i],&ed[i]);
		dfs(st[i],-1,ed[i],i);
	}
	for (int i=1;i<n;i++) ans+=min(Cnt[i],2);
	printf("%d\n",ans);
	while (1==1)
	{
		int p=0,q;//choose edge 
		for (int i=1;i<n;i++)//如果有一条边的一个方向已被确定，那么剩下的那一条路径有唯一的方向
		  if (Cnt[i]==1&&vis[i*2]!=vis[i*2+1]) {p=i;break;} 
		if (!p)//否则在经过次数>=2的边中随便取一条 
		  for (int i=1;i<n;i++)
		    if (Cnt[i]>=2) {p=i;break;}
	    if (!p) break;
		for (int i=0;i<pass[p].size();i++)//choose path
		  if (!sure[pass[p][i]]) {q=pass[p][i];break;}
		int x=st[q],y=ed[q];sure[q]=1;
		for (int i=0;i<vec[q].size();i++)
		{
			int t=vec[q][i];
			if (t/2==p)
			{
				if (vis[t])
				{
				   for (int j=0;j<vec[q].size();j++)
				     vec[q][j]^=1;
				   swap(st[q],ed[q]);
				}
				for (int j=0;j<vec[q].size();j++)
				  Cnt[vec[q][j]/2]--,vis[vec[q][j]]=1;
				break;
			}
		}
	}
	for (int i=1;i<=m;i++) printf("%d %d\n",st[i],ed[i]);
	return 0;
}
 