Problem:
Score : 1500 points

Problem Statement

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.

Sample Input 1

4 3
2 1
3 1
4 1
2 3
3 4
4 2

Sample Output 1

6
2 3
3 4
4 2

If we decide the directions of the walks as above, he gains the happiness of 2 in each walk, and the total happiness will be 6.

Sample Input 2

5 3
1 2
1 3
3 4
3 5
2 4
3 5
1 5

Sample Output 2

6
2 4
3 5
5 1

Sample Input 3

6 4
1 2
2 3
1 4
4 5
4 6
2 4
3 6
5 6
4 5

Sample Output 3

9
2 4
6 3
5 6
4 5

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define psb push_back

using namespace std;

const int N=4005;

typedef long long LL;

int n,m,tot,h[N],e[N],nxt[N],ans,s[N],X[N],Y[N],lca[N],fa[N];

bool vis[N],bz[N][2],Flip[N],Set[N],Dir[N][N];

vector <int> H[N];

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

void dfs(int x)
{
	for (int i=h[x];i;i=nxt[i]) if (e[i]!=fa[x])
	{
		fa[e[i]]=x; dfs(e[i]);
	}
}

void work(int x,int y)
{
	bz[x][y]=1;
	if (vis[x]) return;
	vis[x]=1;
	if (H[x].size()==1) return;
	
	if (!bz[x][y^1])
	{
		int k=0;
		for (;k<H[x].size() && Set[H[x][k]];k++);
		if (k==H[x].size()) return;
		k=H[x][k];
		Set[k]=1; Flip[k]=Dir[k][x]^y^1;
		for (int i=X[k];i!=lca[k];i=fa[i]) work(i,Flip[k]);
		for (int i=Y[k];i!=lca[k];i=fa[i]) work(i,Flip[k]^1);
	}

}

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1,x,y;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		Add(x,y); Add(y,x);
	}
	dfs(1);
	for (int i=0;i<m;i++)
	{
		scanf("%d%d",X+i,Y+i);
		memset(vis,0,sizeof(vis));
		for (int j=X[i];j;j=fa[j]) vis[j]=1;
		for (lca[i]=Y[i];!vis[lca[i]];lca[i]=fa[lca[i]]);
		for (int j=X[i];j!=lca[i];j=fa[j]) H[j].psb(i);
		for (int j=Y[i];j!=lca[i];j=fa[j]) H[j].psb(i),Dir[i][j]=1;
	}
	for (int i=2;i<=n;i++) if (H[i].size()) ans+=((H[i].size()==1)?1:2);
	printf("%d\n",ans);
	memset(vis,0,sizeof(vis));
	for (int i=2;i<=n;i++) if (!vis[i] && H[i].size()) //work(i,0);
	{
		int j=H[i][0];
		Set[j]=1; Flip[j]=0;
		for (int k=X[j];k!=lca[j];k=fa[k]) work(k,0);
		for (int k=Y[j];k!=lca[j];k=fa[k]) work(k,1);
	}
	for (int i=0;i<m;i++) if (Flip[i]) printf("%d %d\n",Y[i],X[i]);else printf("%d %d\n",X[i],Y[i]);
	return 0;
}