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>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int N=2005;
int u[N],v[N],re[N],vis[N],owo[N<<1];
int n,m,a,b,ans;
vector<pii> e[N];
vi p[N],id[N];

int dfs(int u,int fa,int aim,int from){
	if(u==aim)return 1;
	for(auto v:e[u])	
		if(v.fi!=fa&&dfs(v.fi,u,aim,from))
			return p[from].pb(v.se),1;
	return 0;
}

int main(){
	read(n),read(m);
	rep(i,1,n-1){
		read(a),read(b);
		e[a].pb(mp(b,2*i));
		e[b].pb(mp(a,2*i+1));
	}
	rep(i,1,m){
		read(u[i]),read(v[i]);
		dfs(u[i],0,v[i],i);
		for(auto k:p[i])id[k/2].pb(i),re[k/2]++;
	}
	rep(i,1,n-1)ans+=min(2,re[i]);
	for(;;){
		int f=0,g=0;
		rep(i,1,n-1)
			if(re[i]==1&&owo[i*2]!=owo[i*2+1]){f=i;break;}
		rep(i,1,n-1)
			if(re[i]>=2&&!f){f=i;break;}
		if(!f)break;
		for(auto k:id[f])
			if(!vis[k]){g=k;break;}
		vis[g]=1;
		for(auto k:p[g]){
			if(k/2!=f)continue;
			if(owo[k]){
				swap(u[g],v[g]);
				for(auto &j:p[g])j^=1;
			}
			for(auto j:p[g])re[j/2]--,owo[j]=1;
		}
	}
	printf("%d\n",ans);
	rep(i,1,m)printf("%d %d\n",u[i],v[i]);
	return 0;
}