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 <cstdio>
#include <vector>
#include <queue>
using namespace std;
#define N 2050
int n,m,id,i;
int u[N*3],v[N*3],a[N*3],faa[N],dep[N],fa[N*3],in[N],up[N];
bool vis[N*3],viss[N];
vector<int> g[N],son[N*3],S[N];
queue<int> q;
void dfs(int now)
{
	if (vis[now]) return;vis[now]=true;a[now]^=a[fa[now]];
	for (int ii=0;ii<son[now].size();ii++) dfs(son[now][ii]);
}
void gtdp(int now,int lst)
{
	faa[now]=lst;dep[now]=dep[lst]+1;
	for (int ii=0;ii<g[now].size();ii++) if (g[now][ii]!=lst) gtdp(g[now][ii],now);
}
int main()
{
	scanf("%d%d",&n,&m);id=m;
	for (i=1;i<n;i++){
		int x,y;scanf("%d%d",&x,&y);
		g[x].push_back(y);g[y].push_back(x);in[x]++;in[y]++;
	}for (i=1;i<=n;i++) if (in[i]==1) q.push(i);gtdp(1,0);
	for (i=1;i<=m;i++){
		scanf("%d%d",&u[i],&v[i]);S[u[i]].push_back(i);S[v[i]].push_back(i);
		int x=u[i],y=v[i];
		while (x!=y){
			if (dep[x]<dep[y]) swap(x,y);
			up[x]++;x=faa[x];
		}
	}int ans=0;
	for (i=1;i<=n;i++) ans+=min(up[i],2);printf("%d\n",ans);
	while (!q.empty()){
		int psz=q.front();q.pop();int tot=0;
		for (i=0;i<S[psz].size();i++) if (u[S[psz][i]]==psz&&v[S[psz][i]]==psz) dfs(S[psz][i]);
		else if (!fa[S[psz][i]]) tot++;
		while (tot>=2){
			for (i=0;i<S[psz].size();i++) if (!(u[S[psz][i]]==psz&&v[S[psz][i]]==psz)&&!fa[S[psz][i]]) break;
			int ii=i;for (i++;i<S[psz].size();i++) if (!(u[S[psz][i]]==psz&&v[S[psz][i]]==psz)&&!fa[S[psz][i]]) break;
			int uu,vv;
			if (u[S[psz][i]]==psz) a[S[psz][i]]^=1,uu=v[S[psz][i]];else uu=u[S[psz][i]];
			if (u[S[psz][ii]]==psz) vv=v[S[psz][ii]];else a[S[psz][ii]]^=1,vv=u[S[psz][ii]];
			id++;u[id]=uu;v[id]=vv;S[uu].push_back(id);S[vv].push_back(id);fa[S[psz][i]]=id;fa[S[psz][ii]]=id;son[id].push_back(S[psz][i]);son[id].push_back(S[psz][ii]);
			tot-=2;
		}if (tot){
			for (i=0;i<S[psz].size();i++) if (!(u[S[psz][i]]==psz&&v[S[psz][i]]==psz)&&!fa[S[psz][i]]) break;
			int fdc;
			for (int ii=0;ii<g[psz].size();ii++) if (!viss[g[psz][ii]]) fdc=g[psz][ii];
			id++;fa[S[psz][i]]=id;son[id].push_back(S[psz][i]);
			if (u[S[psz][i]]==psz) u[id]=fdc,v[id]=v[S[psz][i]];else u[id]=u[S[psz][i]],v[id]=fdc;
			S[u[id]].push_back(id);S[v[id]].push_back(id);
		}for (int ii=0;ii<g[psz].size();ii++) if (!viss[g[psz][ii]]){
			if (--in[g[psz][ii]]==1) q.push(g[psz][ii]);
		}viss[psz]=true;
	}
	for (i=1;i<=m;i++) if (a[i]) printf("%d %d\n",u[i],v[i]);else printf("%d %d\n",v[i],u[i]);
	return 0;
}