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 FR first
#define SE second

using namespace std;

typedef pair<int,int> pr;

vector <int> e[2005];
int d[2005];

int fa[2005],dep[2005];

void dfs1(int x) {
  for(int i=0;i<e[x].size();i++)
    if (e[x][i]!=fa[x]) {
    	int u=e[x][i];
    	fa[u]=x;dep[u]=dep[x]+1;
    	dfs1(u);
	}
}

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 sum[2005];

void dfs2(int x) {
  for(int i=0;i<e[x].size();i++)
    if (e[x][i]!=fa[x]) {
    	int u=e[x][i];
    	dfs2(u);
    	sum[x]+=sum[u];
	} 
}

int p[2005];
bool col[2005],vis[2005];

pr num[2005],fir[2005];

int main() {
  int n,m;
  scanf("%d%d",&n,&m);
  for(int i=1;i<n;i++) {
  	int x,y;
  	scanf("%d%d",&x,&y);
  	e[x].push_back(y);
  	e[y].push_back(x);
  	d[x]++;d[y]++;
  }
  dfs1(1);
  for(int i=1;i<=m;i++) {
  	int x,y;
  	scanf("%d%d",&x,&y);
  	fir[i]=num[i]=pr(x,y);
  	sum[x]++;sum[y]++;
  	sum[lca(x,y)]-=2;
  }
  dfs2(1);
  int ans=0;
  for(int i=2;i<=n;i++) ans+=min(sum[i],2);
  printf("%d\n",ans);
  for(int i=2;i<=n;i++) {
  	int u=0;
  	for(int j=2;j<=n;j++)
  	  if (d[j]==1&&!vis[j]) {
  	  	  u=j;
  	  	  break;
		}
	vis[u]=1;
	int v=0;
	for(int j=0;j<e[u].size();j++)
	  if (!vis[e[u][j]]) {
	  	v=e[u][j];
	  	break;
	  }
	d[v]--;
	int id1=0,id2=0;
	for(int j=1;j<=m;j++)
	  if ((num[j].FR==u)^(num[j].SE==u)) {
	  	if (!id1) id1=j;
	  	else if (!id2) id2=j;
	  }
	if (!id1&&!id2) continue;
	if (!id2) {
		if (num[id1].FR==u) num[id1].FR=v;
		else num[id1].SE=v; 
		continue;
	}
	if (num[id1].FR==u) {
		if (num[id2].FR==u) {
			num[id1].FR=num[id2].SE;
			col[id2]^=1;
		}
		else num[id1].FR=num[id2].FR;
	}
	else {
		if (num[id2].SE==u) {
			num[id1].SE=num[id2].FR;
			col[id2]^=1;
		}
		else num[id1].SE=num[id2].SE;
	}
	num[id2].FR=num[id2].SE;
	p[id2]=id1;
	for(int j=1;j<=m;j++)
	  if ((num[j].FR==u)^(num[j].SE==u)) {
	  	if (num[j].FR==u) num[j].FR=v;
	  	else num[j].SE=v;
	  }
  }
  for(int i=1;i<=m;i++)
    if (p[i]) col[i]^=col[p[i]];
  for(int i=1;i<=m;i++) {
  	if (col[i]) swap(fir[i].FR,fir[i].SE);
  	printf("%d %d\n",fir[i].FR,fir[i].SE);
  }
  return 0;
}