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;
typedef long long ll;
const int N=2005;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,m,ans,cnt,b[N],fa[N],depth[N],c[N],h[N*3],fr[N*3];
struct node{int u,v;}a[N*3];
vector<int> e[N];
unordered_set<int> g[N];
void work(int x,int fath){
	fa[x]=fath;depth[x]=depth[fath]+1;
	for (auto y : e[x])
	if (y!=fath) work(y,x);
	b[++cnt]=x;
}
void add(int u,int v){
	a[++cnt]=(node){u,v};
	if (u==v) return;
	g[u].insert(cnt);g[v].insert(cnt);
}
void calc(int x,int i,int j){
	int u=a[i].u+a[i].v-x,v=a[j].u+a[j].v-x;
	g[u].erase(i);g[v].erase(j);
	add(u,v);
	fr[i]=fr[j]=cnt;
	h[i]=(a[i].v==u);h[j]=(a[j].u==v);
}
void chk(int x,int y){
	while (x!=y){
		if (depth[x]<depth[y]) swap(x,y);
		c[x]++;x=fa[x];
	}
}
int main(){
	n=read();m=read();
	for (int i=1;i<n;i++){
		int u=read(),v=read();
		e[u].push_back(v);
		e[v].push_back(u);
	}
	work(1,0);cnt=0;
	for (int i=1;i<=m;i++){
		int u=read(),v=read();
		add(u,v);chk(u,v);
	}
	for (int i=1;i<=n;i++){
		int x=b[i],las=0;
		for (auto now : g[x])
		if (!las) las=now;
		else {calc(x,las,now);las=0;}
		if (las){
			int y=a[las].u+a[las].v-x;
			g[y].erase(las);
			add(fa[x],y);
			fr[las]=cnt;h[las]=(a[las].v==x);
		}
	}
	for (int i=2;i<=n;i++) ans+=min(c[i],2);
	printf("%d\n",ans);
	for(int i=cnt;i;i--) h[i]^=h[fr[i]];
	for (int i=1;i<=m;i++)
	if (!h[i]) printf("%d %d\n",a[i].u,a[i].v);
	else printf("%d %d\n",a[i].v,a[i].u);
	return 0;
}