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<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
const int N=2005*2;
int n,m;
struct qq
{
	int x,y,last;
}e[N];
int num,last[N];
void init (int x,int y)
{
	e[num].x=x;e[num].y=y;
	e[num].last=last[x];
	last[x]=num;
	num++;
}
int x[N],y[N];
vector<int> path[N],vec[N];
bool vis[N],used[N];//用过没有 
int sum[N];
bool dfs (int x,int fa,int t,int p)
{
	if (x==t) return true;
	for (int u=last[x];u!=-1;u=e[u].last)
	{
		int y=e[u].y;
		if (y==fa) continue;
		if (dfs(y,x,t,p))
		{
			path[p].push_back(u);
			sum[u>>1]++;
			vec[u>>1].push_back(p);
			return true;
		}
	}
	return false;
}
int main()
{
	num=0;memset(last,-1,sizeof(last));
	memset(used,false,sizeof(used));
	memset(vis,false,sizeof(vis));
	scanf("%d%d",&n,&m);
	for (int u=1;u<n;u++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		init(x,y);init(y,x);
	}
	for (int u=1;u<=m;u++)
	{
		scanf("%d%d",&x[u],&y[u]);
		dfs(x[u],0,y[u],u);
	}
	int ans=0;
	for (int u=0;u<n-1;u++)	ans=ans+min(sum[u],2);
	while (true)
	{
		int id=-1;
		for (int u=0;u<n-1;u++)
		{
			if (sum[u]==1&&(vis[u<<1]^vis[u<<1|1])==1)
			{
				id=vis[u<<1]==1?u<<1|1:u<<1;
				break;
			}
		}
	
		if (id==-1)
		{
			for (int u=0;u<n-1;u++)
			{
				if (sum[u]>0)//随便决策一下 
				{
					id=(u<<1);
					break;
				}
			}
			if (id==-1) break;
		}	
		int p=(id>>1);
		while (used[vec[p].back()]) vec[p].pop_back();
		int t=vec[p].back();vec[p].pop_back();
		used[t]=true;
		bool rev=false;
		for (int u=0;u<path[t].size();u++)
		{
			if ((path[t][u]^id)==1)
			{
				rev=true;
				break;
			}
		}
		if (rev==true)
		{
			for (int u=0;u<path[t].size();u++) path[t][u]^=1;
			swap(x[t],y[t]);
		}
		for (int u=0;u<path[t].size();u++) 
		{
			sum[path[t][u]>>1]--;
			vis[path[t][u]]=true;
		}
	}
	printf("%d\n",ans);
	for (int u=1;u<=m;u++)printf("%d %d\n",x[u],y[u]);
	return 0;
}