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>
using namespace std;
const int N=4050;
struct data{
  int x,y;
}w[N];
int head[N],num[N],nn[N][2],cnt=1;
#define link(u,v) (nn[++cnt][1]=v,nn[cnt][0]=head[u],head[u]=cnt)
bool vis[N],val[N];
vector<int>ed[N],st[N];
vector<int>::iterator it;
int dfs(int x,int id,int fa,int t){
  if (x==t) return 1;
  for (int i=head[x],y; i; i=nn[i][0])
    if ((y=nn[i][1])!=fa&&dfs(y,id,x,t)){
	ed[id].push_back(i);
	st[i>>1].push_back(id);
	return 1;
      }
  return 0;
}
int main(){
  int n,m,u,v; cin>>n>>m;
  for (int i=2; i<=n; ++i)
    cin>>u>>v,link(u,v),link(v,u);
  for (int i=1; i<=m; ++i){
    cin>>w[i].x>>w[i].y;
    dfs(w[i].x,i,0,w[i].y);
  }
  int ans=0;
  for (int i=1; i<n; ++i)
    num[i]=st[i].size(),ans+=min(2,num[i]);
  for (;;){
    int x;
    for (x=1; x<n; ++x){
      if (num[x]==1&&vis[x<<1]!=vis[x<<1|1]) break;
    }
    if (x==n){
      for (x=1; x<n&&!num[x]; ++x);
      if (x==n) break;
    }
    while(val[st[x].back()]) st[x].pop_back();
    int p=st[x].back(),rev=0;
    val[p]=1;
    for (it=ed[p].begin(); it!=ed[p].end(); ++it)
      if ((*it)/2==x){
	if (vis[*it])
	  rev=1,swap(w[p].x,w[p].y);
	break;
      }
    for (it=ed[p].begin(); it!=ed[p].end(); ++it)
      vis[*it^rev]=1,--num[*it>>1];
  }
  cout<<ans<<endl;
  for (int i=1; i<=m; ++i)
    cout<<w[i].x<<" "<<w[i].y<<endl;
  return 0;
}