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>
#define ll long long
#define pb emplace_back
#define mr make_pair
#define N 2010
using namespace std;
int m,n,x,y,d[N],f[N],c[N],nc,g[N],ne,p[N],q[N],ans;
bool vi[N];
vector<int> a[N],s[N];
struct Info{int x,y;bool ex;}e[N*2];
void jb(int x,int y){a[x].pb(y);}
void dfs(int x,int fa,int dep){
	f[x]=fa;d[x]=dep;
	for (auto y:a[x])if (y!=fa)dfs(y,x,dep+1);
	c[++nc]=x;vi[x]=1;
}
void get(int x,int y){
	while (x!=y){
		if(d[x]<d[y])swap(x,y);
		g[x]++;x=f[x];
	}
}
void g1(int x,int y){while (x!=y){if(d[x]<d[y])swap(x,y);vi[x]=0;x=f[x];}}
int lca(int x,int y){while (x!=y){if(d[x]<d[y])swap(x,y);x=f[x];}return x;}
int main(){
	cin>>n>>m;ne=m;
	for (int i=1;i<n;i++){cin>>x>>y;jb(x,y);jb(y,x);}
	dfs(1,0,1);
	for (int i=1;i<=m;i++){cin>>e[i].x>>e[i].y;get(e[i].x,e[i].y);e[i].ex=1;}
	for (int i=1;i<=n;i++){
		ans+=min(g[c[i]],2);
		if (vi[c[i]]){
			if (g[c[i]]>1){
				for (int j=1;j<=ne;j++){
					if (e[j].ex&&e[j].x!=e[j].y&&(e[j].x==c[i]||e[j].y==c[i])){
						if (p[i]){q[i]=j;break;}else p[i]=j;
					}
				}
				int x=lca(e[p[i]].x,e[p[i]].y),y=lca(e[q[i]].x,e[q[i]].y);
				if (x==y){
					if (e[p[i]].x!=c[i])x=e[p[i]].x;else x=e[p[i]].y;
					if (e[q[i]].x!=c[i])y=e[q[i]].x;else y=e[q[i]].y;
					x=lca(x,y);				
				}else {if (d[x]<d[y]) swap(x,y);}
				g1(c[i],x);
				e[p[i]].ex=0;e[q[i]].ex=0;
				ne++;e[ne].ex=1;
				if (e[p[i]].x==c[i])e[ne].x=e[p[i]].y;else e[ne].x=e[p[i]].x;
				if (e[q[i]].x==c[i])e[ne].y=e[q[i]].y;else e[ne].y=e[q[i]].x;
			}			
		}
		for (int j=1;j<=ne;j++){
			if (e[j].ex&&e[j].x!=e[j].y&&(e[j].x==c[i]||e[j].y==c[i])){
				if (e[j].x==c[i])e[j].x=f[c[i]];else e[j].y=f[c[i]];
				s[i].pb(j);
			}
		}
	}
	for (int i=n;i>=1;i--){
//		cout<<i<<' '<<p[i]<<' '<<q[i]<<' '<<c[i]<<endl;
//		for (int j=1;j<=ne;j++)cout<<e[j].x<<' '<<e[j].y<<endl;
		if (p[i]){
			if (e[ne].x==e[p[i]].y||e[ne].y==e[p[i]].x)swap(e[p[i]].x,e[p[i]].y);
			if (e[ne].x==e[q[i]].y||e[ne].y==e[q[i]].x)swap(e[q[i]].x,e[q[i]].y);
			if (e[ne].x==e[ne].y&&e[p[i]].x==e[q[i]].x)swap(e[q[i]].x,e[q[i]].y);
			ne--;
		}
		for (auto j:s[i]){
			if (e[j].x==f[c[i]])e[j].x=c[i];else e[j].y=c[i];
		}
	}
	cout<<ans<<endl;
	for (int i=1;i<=m;i++)cout<<e[i].x<<' '<<e[i].y<<endl;
	return 0;
}