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>
#define Maxn 4007
using namespace std;
int n,m;
int last[Maxn],pre[Maxn],other[Maxn],cnt=1;
int a[Maxn],b[Maxn];
vector<int> p[Maxn],cover[Maxn];
bool flag[Maxn],vis[Maxn];
int tot[Maxn];
void insert(int u,int v)
{
	other[++cnt]=v,pre[cnt]=last[u],last[u]=cnt;
}
bool dfs(int u,int fa,int g,int s)
{
	if (u==g) return true;
	for (int q=last[u];q;q=pre[q])
	{
		int v=other[q];
		if (v!=fa)
		{
			if (dfs(v,u,g,s))
			{
				p[s].push_back(q);
				return true;
			}
		}
	}
	return false;
}
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		insert(u,v),insert(v,u);
	}
	for (int i=1;i<=m;i++)
	{
		scanf("%d%d",&a[i],&b[i]);
		dfs(a[i],-1,b[i],i);
		for (int j=0;j<(int)p[i].size();j++)
		{
			++tot[p[i][j]/2];
			cover[p[i][j]/2].push_back(i);
		}
	}
	int ans=0;
	for (int i=1;i<n;i++)
		ans+=min(tot[i],2);
	while (true)
	{
		int pos=0;
		for (int i=1;i<n;i++)
			if (tot[i]==1&&(flag[i*2]^flag[i*2+1]))
			{
				if (flag[i*2]) pos=i*2+1; else pos=i*2;
				break;
			}
		if (pos==0)
		{
			for (int i=1;i<n;i++)
				if (tot[i]>0) pos=i*2;
		}
		if (pos==0) break;
		while (vis[cover[pos/2].back()]) cover[pos/2].pop_back();
		int now=cover[pos/2].back();
		vis[now]=true;
		bool check=false;
		for (int j=0;j<(int)p[now].size();j++)
			if (p[now][j]/2==pos/2)
			{
				if (p[now][j]^pos) 
				{
					check=true;
					break;
				}
			}
		if (check)
		{
			swap(a[now],b[now]);
			for (int j=0;j<(int)p[now].size();j++)
				p[now][j]^=1;
		}
		for (int j=0;j<(int)p[now].size();j++)
			flag[p[now][j]]=true,--tot[p[now][j]/2];
	}
	printf("%d\n",ans);
	for (int i=1;i<=m;i++)
		printf("%d %d\n",a[i],b[i]);
	return 0;
}