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:
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
#define N 2500
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
struct edge{int to,lt;}e[N<<1];
int in[N],cnt,fa[N],dep[N],n,m,d[N];
void add(int x,int y)
{
	e[++cnt].to=y; e[cnt].lt=in[x]; in[x]=cnt; d[x]++;
	e[++cnt].to=x; e[cnt].lt=in[y]; in[y]=cnt; d[y]++;
}
void dfs(int x,int fr)
{
	dep[x]=dep[fr]+1; fa[x]=fr;
	for(int i=in[x];i;i=e[i].lt)
		if(e[i].to!=fr)	dfs(e[i].to,x);
}
int LCA(int x,int y)
{
	if(dep[x]<dep[y])	swap(x,y);
	while(dep[x]>dep[y]) x=fa[x];
	while(x!=y)	x=fa[x],y=fa[y];
	return x;
}
int s[N]; bool vis[N],tag[N];
void calc(int x)
{
	for(int i=in[x];i;i=e[i].lt) if(e[i].to!=fa[x]) calc(e[i].to),s[x]+=s[e[i].to];
}
int a[N],b[N],ta[N],tb[N],p[N];
int main()
{
	n=read(),m=read(); for(int i=1;i<n;i++)	add(read(),read());
	dfs(1,0); for(int i=1;i<=m;i++)
		ta[i]=a[i]=read(),tb[i]=b[i]=read(),
		s[a[i]]++,s[b[i]]++,s[LCA(a[i],b[i])]-=2;
	calc(1); int T=0; for(int i=2;i<=n;i++) T+=min(2,s[i]); printf("%d\n",T);
	for(int t=2;t<=n;t++)
	{
		int x=0;
		for(int j=2;j<=n;j++)	if(d[j]==1&&!vis[j]){x=j; break;} vis[x]=1; 
		int fx=fa[x],i1=0,i2=0; d[fx]--;
		for(int i=1;i<=m;i++)	if((ta[i]==x)^(tb[i]==x))
			if(!i1)	i1=i; else{i2=i; break;}
		if(!i1&&!i2)	continue;
		if(!i2){if(ta[i1]==x) ta[i1]=fx; else tb[i1]=fx; continue;}
		if(ta[i1]==x)
		{
			if(ta[i2]==x) ta[i1]=tb[i2],tag[i2]^=1;
			else ta[i1]=ta[i2];
		}
		else
		{
			if(tb[i2]==x) tb[i1]=ta[i2],tag[i2]^=1;
			else	tb[i1]=tb[i2];
		}
		ta[i2]=tb[i2]; p[i2]=i1;
		for(int i=i2+1;i<=m;i++) if((ta[i]==x)^(tb[i]==x))
			if(ta[i]==x) ta[i]=fx;
			else	tb[i]=fx;
	}
	for(int i=1;i<=m;i++)	if(p[i]) tag[i]^=tag[p[i]];
	for(int i=1;i<=m;i++)
	{
		if(tag[i])	printf("%d %d\n",b[i],a[i]);
		else	printf("%d %d\n",a[i],b[i]);
	}
	return 0;
}