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<utility>
#include<algorithm>
#include<vector>
using namespace std;
typedef pair<int,int> pii;
vector<int> e[2100];
int n,m,x,y,deg[2100],sum[2100];
int fa[2100],dep[2100],p[2100];
bool col[2100],vis[2100];
pii num[2100],q[2100];

void dfs1(int u){
	for (int i=0;i<(int)e[u].size();i++)
		if (e[u][i]!=fa[u]){
			int v=e[u][i];
			fa[v]=u; dep[v]=dep[u]+1;
			dfs1(v);
		}
}

void dfs2(int u){
	int v;
	for (int i=0;i<(int)e[u].size();i++)
		if (e[u][i]!=fa[u]){
			v=e[u][i]; dfs2(v);
			sum[u]+=sum[v];
		} 
}

int lca(int x,int y){
	if (dep[x]<dep[y]) swap(x,y);
	while (dep[x]>dep[y]) x=fa[x];
	while (x!=y) x=fa[x],y=fa[y];
	return x;
}

int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
		deg[x]++; deg[y]++;
	}
	dfs1(1);
	for (int i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		q[i]=pii(x,y); num[i]=pii(x,y);
		sum[x]++; sum[y]++;
		sum[lca(x,y)]-=2;
	}
	dfs2(1);
	int ans=0;
	for (int i=2;i<=n;i++) ans+=min(sum[i],2);
	printf("%d\n",ans);
	for (int i=2;i<=n;i++){
		int u=0;
		for (int j=2;j<=n;j++)
			if (deg[j]==1&&!vis[j]){
				u=j;
				break;
			}
		vis[u]=true;
		int v=0;
		for (int j=0;j<(int)e[u].size();j++)
			if (!vis[e[u][j]]){
				v=e[u][j];
				break;
			}
		deg[v]--;
		
		int a=0,b=0;
		for (int j=1;j<=m;j++)
			if ((num[j].first==u)^(num[j].second==u)){
				if (!a) a=j;
				else if (!b) b=j;
				else break;
			}
		if (!a&&!b) continue;
		if (!b){
			if (num[a].first==u) num[a].first=v;
			else num[a].second=v; 
			continue;
		}
		if (num[a].first==u){
			if (num[b].first==u){
				num[a].first=num[b].second;
				col[b]^=1;
			}
			else num[a].first=num[b].first;
		}
		else{
			if (num[b].second==u){
				num[a].second=num[b].first;
				col[b]^=1;
			}
			else num[a].second=num[b].second;
		}
		num[b].first=num[b].second;
		p[b]=a;
		for (int j=1;j<=m;j++)
			if ((num[j].first==u)^(num[j].second==u)){
				if (num[j].first==u) num[j].first=v;
				else num[j].second=v;
			}
	}
	for (int i=1;i<=m;i++)
		if (p[i]) col[i]^=col[p[i]];
		
	for (int i=1;i<=m;i++){
		if (col[i]) swap(q[i].first,q[i].second);
		printf("%d %d\n",q[i].first,q[i].second);
	}
	return 0;
}