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=2005;
vector<int> a[N][N];
int vis[N],u[N],v[N],tot,ne[N*2],fi[N],zz[N*2],flag[N][N],out[N],l,in[N];
int x[N],y[N],cnt[N][N],n,m,ans,fa[N];
void jb(int x,int y){
	ne[++tot]=fi[x];
	fi[x]=tot;
	zz[tot]=y;
	flag[x][y]=1;
}
void dfs(int x,int y){
	fa[x]=y;in[x]=++l;
	for (int i=fi[x];i;i=ne[i])
		if (zz[i]!=y)dfs(zz[i],x);
	out[x]=++l;
}
int pd(int x,int y){
	return in[x]<=in[y]&&out[x]>=out[y];
}
void fg(int x,int y){
	while (!pd(x,y)){
		ans+=flag[x][fa[x]];
		flag[x][fa[x]]=0;
		cnt[x][fa[x]]--;
		cnt[fa[x]][x]--;
		x=fa[x];
	}
	while (x!=y){
		ans+=flag[fa[y]][y];
		flag[fa[y]][y]=0;
		cnt[y][fa[y]]--;
		cnt[fa[y]][y]--;
		y=fa[y];
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<n;i++){
		scanf("%d%d",&x[i],&y[i]);
		jb(x[i],y[i]);jb(y[i],x[i]);
	}
	dfs(1,0);
	for (int i=1;i<=m;i++){
		scanf("%d%d",&u[i],&v[i]);
		int x=u[i],y=v[i];
		while (!pd(x,y)){
			a[x][fa[x]].push_back(i);
			a[fa[x]][x].push_back(-i);
			x=fa[x];
		}
		while (x!=y){
			a[y][fa[y]].push_back(-i);
			a[fa[y]][y].push_back(i);
			y=fa[y];
		}
	}
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)cnt[i][j]=a[i][j].size();
	while (1){
		int p=0;
		for (int i=1;i<n;i++)
			if (cnt[x[i]][y[i]]==1&&flag[x[i]][y[i]]!=flag[y[i]][x[i]]){
				p=i;
				break;
			}
		if (!p){
			for (int i=1;i<n;i++)
				if (cnt[x[i]][y[i]]>=2){
					p=i;
					break;
				}
		}
		if (!p)for (int i=1;i<n;i++)
			if (cnt[x[i]][y[i]]>=1){
				p=i;
				break;
			}
		if (!p)break;
		if (!flag[x[p]][y[p]])swap(x[p],y[p]);
		for (int i=0;i<a[x[p]][y[p]].size();i++)
			if (!vis[abs(a[x[p]][y[p]][i])]){
				vis[abs(a[x[p]][y[p]][i])]=1;
				int now=abs(a[x[p]][y[p]][i]);
				if (a[x[p]][y[p]][i]<0)swap(u[now],v[now]);
				fg(u[now],v[now]);
				break;
			}
	}
	printf("%d\n",ans);
	for (int i=1;i<=m;i++)printf("%d %d\n",u[i],v[i]);
}