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>
using namespace std;
#define N 4001
int he[N],ne[(N<<1)-1],e[(N<<1)-1],cnt=1;
int St[N],Ed[N],d[N],w[(N<<1)-1],vis[N],n,m;
vector<int>b[(N<<1)-1],c[(N<<1)-1];

void add(int u,int v){
	ne[++cnt]=he[u];
	he[u]=cnt;
	e[cnt]=v;
}

bool dfs(int x,int fa,int Ed,vector<int>&b){
	if(x==Ed)return 1;
	for(int i=he[x];i;i=ne[i]){
		if(e[i]!=fa){
			b.push_back(i);
			if(dfs(e[i],x,Ed,b)) return 1;
			b.pop_back();
		}
	}
	return 0;
}

int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<n;++i){
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y);
		add(y,x);
	}
	for(int i=0;i<m;++i){
		scanf("%d%d",&St[i],&Ed[i]);
		dfs(St[i],0,Ed[i],b[i]);
		for (int j=0;j<b[i].size();j++){
			c[b[i][j]/2].push_back(i);
			++d[b[i][j]/2];
		}
	}
	int ans=0;
	for(int i=1;i<n;++i)ans+=min(2,d[i]);
	printf("%d\n",ans);
	for(;;){
		int p=0,q=-1;
		for(int i=1;i<n;++i){
			if(d[i]==1&&w[i*2]!=w[i*2+1]){
				p=i;
				break;
			}
		}
		if(p==0){
			for(int i=1;i<n;++i){
				if(d[i]>=2){
					p=i;
					break;
				}
			}
		}
		if(p==0)break;
		for(int i=0;i<c[p].size();++i){
			if(!vis[c[p][i]]){
				q=c[p][i];
				break;
			}
		}
		vis[q]=1;
		for (int i=0;i<b[q].size();++i){
			if (b[q][i]/2==p){
				if (w[b[q][i]]){
					swap(St[q],Ed[q]);
					for(int j=0;j<b[q].size();++j)
					b[q][j]^=1;
				}
				for (int j=0;j<b[q].size();j++){
					--d[b[q][j]/2];
					w[b[q][j]]=1;
				}
				break;
			}
		}
	}
	for(int i=0;i<m;++i)
	printf("%d %d\n",St[i],Ed[i]);
	return 0;
}