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<iostream>
#include<cstdio>
#include<queue>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=100100;
int n,m,T,u,v,ans,du[N],d[N],t[N],g[N],b[N],c[N],fa[N];
struct lin{int x,y;}p[N],f[N];
vector<int>ad[N],ve;
queue<int>q;
void add(int u,int v){ad[u].push_back(v);du[v]++;}
void dfs(int u){for(int v:ad[u])if(v!=fa[u]) fa[v]=u,d[v]=d[u]+1,dfs(v);}
void gtp(int x,int y){
	if(d[x]<d[y]) swap(x,y);
	for(;d[x]>d[y];x=fa[x]) c[x]++;
	for(;x!=y;x=fa[x],y=fa[y]) c[x]++,c[y]++; 
}
int main(){
	//freopen("1.in","r",stdin);
	scanf("%d%d",&n,&m);T=m;
	FOR(i,2,n) scanf("%d%d",&u,&v),add(u,v),add(v,u);
	FOR(i,1,m) scanf("%d%d",&p[i].x,&p[i].y),f[i]=p[i];
	dfs(1);
	FOR(i,1,m) gtp(p[i].x,p[i].y);
	FOR(i,1,n) ans+=min(2,c[i]),d[i]=0;
	FOR(i,1,n)if(du[i]==1) q.push(i);
	while(q.size()){
		int u=q.front(),p0=0,p1=0,w;
		q.pop();ve.clear();
		for(int v:ad[u])if(du[v]){
			w=v;du[u]--;
			if(--du[w]==1) q.push(w);
		}
		FOR(i,1,T)if(!b[i]){
			b[i]=f[i].x==u&&f[i].y==u; 
			if(!(f[i].x==u^f[i].y==u)) continue;
			ve.push_back(i);
		}
		//cout<<u<<'\n';
		//FOR(i,1,T)if(!b[i]) cout<<i<<':'<<f[i].x<<' '<<f[i].y<<'\n'; 
		while(ve.size()>1){
			int v0=ve.back(),r0=f[v0].y==u;ve.pop_back();
			int v1=ve.back(),r1=f[v1].y==u;ve.pop_back();
			b[v0]=b[v1]=1;
			f[++T]=(lin){r0?f[v0].x:f[v0].y,r1?f[v1].x:f[v1].y};
			g[v0]=g[v1]=T;d[v0]=!r0,d[v1]=r1;
		}
		if(ve.size()==1){
			int v=ve[0];
			if(f[v].x==u) f[v].x=w;
			if(f[v].y==u) f[v].y=w;
		}
	}
	ROF(i,T,1) t[i]=d[i]^t[g[i]];
	cout<<ans<<'\n';
	FOR(i,1,m){
		if(t[i]) swap(p[i].x,p[i].y);
		cout<<p[i].x<<' '<<p[i].y<<'\n';
	}
}