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
#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define db double
#define pint pair<int,int>
#define mk(x,y) make_pair(x,y)
#define fir first
#define sec second
#define Rep(x,y,z) for(int x=y;x<=z;x++)
#define Red(x,y,z) for(int x=y;x>=z;x--)
using namespace std;
const int MAXN=2e3+5;
char buf[1<<12],*p1=buf,*p2=buf,nc;int ny;
inline char gc() {return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<12,stdin),p1==p2)?EOF:*p1++;}
//inline char gc(){return getchar();}
inline int read(){
	int x=0;for(ny=1;nc=gc(),(nc<48||nc>57)&&nc!=EOF;)if(nc==45)ny=-1;if(nc<0)return nc;
	for(x=nc-48;nc=gc(),47<nc&&nc<58&&nc!=EOF;x=(x<<3)+(x<<1)+(nc^48));return x*ny;
}int n,m,d[MAXN],dep[MAXN],fa[MAXN],Sum[MAXN],e[2][MAXN],ed[2][MAXN],vis[MAXN],p[MAXN],col[MAXN];vector<int>G[MAXN];
void dfs(int x,int f){fa[x]=f,dep[x]=dep[f]+1;for(int i=0,len=G[x].size(),y;i<len;i++)if((y=G[x][i])!=f)dfs(y,x);}
void dfs1(int x){for(int i=0,len=G[x].size(),y;i<len;i++)if((y=G[x][i])!=fa[x])dfs1(y),Sum[x]+=Sum[y];}
inline int lca(int x,int y){for(dep[x]<dep[y]?swap(x,y),0:0;dep[x]>dep[y];x=fa[x]);for(;x!=y;x=fa[x],y=fa[y]);return x;}
int main(){
	n=read(),m=read();
	Rep(i,1,n-1){int x=read(),y=read();G[x].push_back(y),G[y].push_back(x),d[x]++,d[y]++;}
	dfs(1,0);Rep(i,1,m)e[0][i]=ed[0][i]=read(),e[1][i]=ed[1][i]=read(),Sum[e[0][i]]++,Sum[e[1][i]]++,Sum[lca(e[0][i],e[1][i])]-=2;dfs1(1);
	int ans=0;Rep(i,2,n)ans+=min(Sum[i],2);cout<<ans<<'\n';
	Rep(i,1,n-1){
		int u=0,v=0;
		Rep(j,2,n)if(!vis[j]&&d[j]==1){u=j;break;}vis[u]=1;
		for(int j=0,len=G[u].size(),y;j<len;j++)if(!vis[y=G[u][j]]){v=y;break;}d[v]--;
		int id1=0,id2=0;
		Rep(j,1,m)if((ed[0][j]==u)^(ed[1][j]==u))
			if(!id1)id1=j;else if(!id2)id2=j;
		if(!id1&&!id2)continue;
		if(!id2){ed[ed[1][id1]==u][id1]=v;continue;}
		if(ed[0][id1]==u){if(ed[0][id2]==u)ed[0][id1]=ed[1][id2],col[id2]^=1;else ed[0][id1]=ed[0][id2];}
		else if(ed[1][id1]==u){if(ed[1][id2]==u)ed[1][id1]=ed[0][id2],col[id2]^=1;else ed[1][id1]=ed[1][id2];}
		ed[0][id2]=ed[1][id2],p[id2]=id1;
		Rep(j,1,m)if((ed[0][j]==u)^(ed[1][j]==u))ed[ed[1][j]==u][j]=v;
	}Rep(i,1,m)if(p[i])col[i]^=col[p[i]];
	Rep(i,1,m){
		if(col[i])swap(e[0][i],e[1][i]);
		cout<<e[0][i]<<" "<<e[1][i]<<'\n';
	}
	return 0;
}


