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<cstring>
#define M 4004
#define N 2002
int A[N],B[N*N],C[N*N],D[N*N],E[N*N],F[N*N],a[N],b[M],c[M],d[N],f[N],m,n,s,t;
inline void swap(int&u,int&v){s=u,u=v,v=s;}
inline void add(int u,int v,int x,int y){d[v]<d[u]?swap(u,v),0:0,C[++t]=u,D[t]=v,E[t]=x,F[t]=y,u!=v?B[t]=A[u],A[u]=t:0;}
void dfn(int u)
{
    for(int i=a[u];i;i=b[i])if(c[i]!=f[u])f[c[i]]=u,dfn(c[i]);
    d[u]=++t;
}
void dfs(int u)
{
    for(int i=a[u];i;i=b[i])if(c[i]!=f[u])dfs(c[i]);
    for(;B[A[u]];A[u]=B[B[A[u]]])add(D[A[u]],D[B[A[u]]],A[u],B[A[u]]);
    if(A[u])add(f[u],D[A[u]],A[u],0);
}
int i,u,v;
int main()
{
    for(scanf("%d%d",&n,&m),i=t=1;i<n;i++)scanf("%d%d",&u,&v),b[++t]=a[u],c[a[u]=t]=v,b[++t]=a[v],c[a[v]=t]=u;
    for(t=0,dfn(1),i=1,t=0;i<=m;i++)scanf("%d%d",&u,&v),add(u,v,0,0);
    for(dfs(1),i=t;m<i;i--)D[E[i]]==D[i]?swap(C[F[i]],D[F[i]]):swap(C[E[i]],D[E[i]]);
    for(memset(b+1,0,n<<2),memset(c+1,0,n<<2),i=1,s=0;i<=m;i++)for(u=C[i],v=D[i];u!=v;)d[u]<d[v]?s+=!b[u],b[u]=1,u=f[u]:(s+=!c[v],c[v]=1,v=f[v]);
    for(printf("%d\n",s),i=1;i<=m;i++)printf("%d %d\n",C[i],D[i]);
    return 0;
}