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 y1 dmytxdy
#define pb push_back
#define fi first
#define se second
#define mp make_pair

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef long double ld;

template <typename T> bool chkmin(T &x,T y){return x>y?x=y,1:0;}
template <typename T> bool chkmax(T &x,T y){return x<y?x=y,1:0;}

int readint(){
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

int n,m,tot;
int v[4005],nxt[4005],h[2005],dep[2005],vis[2005],f[2005],a[2005],b[2005],hv[2005][2];
set<pii> vec[2005];

void addedge(int x,int y){
	v[++tot]=y; nxt[tot]=h[x]; h[x]=tot;
	v[++tot]=x; nxt[tot]=h[y]; h[y]=tot;
}

void dfs(int u){
	for(int p=h[u];p;p=nxt[p]){
		if(v[p]==f[u]) continue;
		f[v[p]]=u,dep[v[p]]=dep[u]+1;
		dfs(v[p]);
	}
}

void addtag(int x,int y,int id){
	int tx=0,ty=1;
	while(x!=y){
		if(dep[x]<dep[y]) swap(x,y),swap(tx,ty);
		vec[x].insert(mp(id,tx));
		x=f[x];
	}
}

void change(int id,int c){
	int x=a[id],y=b[id],tx=0,ty=1;
	vector<pii> tmp(0);
	while(x!=y){
		if(dep[x]<dep[y]) swap(x,y),swap(tx,ty);
		vec[x].erase(mp(id,tx));
		hv[x][tx^c]=1;
		if(vec[x].size()==1&&!hv[x][tx^c^1]){
			if(vis[vec[x].begin()->fi]==-1){
				vis[vec[x].begin()->fi]=vec[x].begin()->se^tx^c^1;
				tmp.pb(mp(vec[x].begin()->fi,vec[x].begin()->se^tx^c^1));
			}
			else assert(vis[vec[x].begin()->fi]==(vec[x].begin()->se^tx^c^1));
		}
		x=f[x];
	}
	for(auto r:tmp) change(r.fi,r.se);
}

int main(){
	n=readint(); m=readint();
	for(int i=1;i<n;i++) addedge(readint(),readint());
	dep[1]=1; dfs(1);
	for(int i=1;i<=m;i++){
		a[i]=readint(); b[i]=readint();
		addtag(a[i],b[i],i);
	}
	int ans=0;
	for(int i=2;i<=n;i++) ans+=min((int)vec[i].size(),2);
	printf("%d\n",ans);
	memset(vis,-1,sizeof(vis));
	for(int i=m;i>=1;i--){
		if(vis[i]>=0) continue;
		vis[i]=0;
		change(i,0);
	}
	for(int i=1;i<=m;i++) printf("%d %d\n",vis[i]?b[i]:a[i],vis[i]?a[i]:b[i]);
	return 0;
}