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<algorithm>/*{{{*/
#include<cctype>
#include<cassert>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<ctime>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<vector>
using namespace std;
typedef long long lld;
typedef long double lf;
typedef unsigned long long uld;
typedef pair<int,int> pii;
#define fi first
#define se second
#define pb push_back
#define mk make_pair
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define ROF(i,a,b) for(int i=(a);i>=(b);--i)
namespace RA{
    int r(int p){return 1ll*rand()*rand()%p;}
    int r(int L,int R){return r(R-L+1)+L;}
}/*}}}*/
/******************heading******************/
const int N=2030;

int n,m;
vector<int> g[N];
vector<pii> path[N],g_lim[N];
int dep[N],fa[N];
int u[N],v[N],w[N];

void dfs(int u,int p){
    dep[u]=dep[p]+1,fa[u]=p;
    for(int v:g[u])if(v!=p)dfs(v,u);
}
int lca(int x,int y){
    while(x!=y){
        if(dep[x]<dep[y])swap(x,y);
        x=fa[x];
    }
    return x;
}
int pass[N][N],ans;
int choose[N][2];//这个点选择的两条链
void solve(int u,int p){
    choose[u][0]=choose[u][1]=-1;
    FOR(i,1,m)pass[u][i]=-1;
    for(int v:g[u])if(v!=p){
        solve(v,u);
        FOR(i,1,m)if(pass[v][i]!=-1)pass[u][i]=pass[v][i];
    }
    for(pii p:path[u])pass[u][p.fi]=p.se;//继承的同时，ban掉一些以u为lca的边
    int cnt=0;//有多少边经过u的父边
    FOR(i,1,m)cnt+=pass[u][i]!=-1;
    if(cnt<2){
        ans+=cnt;
    }else {
        ans+=2;
        for(int v:g[u])if(v!=p){
            //printf("\033[33m%d,%d\033[0m\n",choose[v][0],choose[v][1]);
            if(choose[v][0]!=-1&&choose[v][1]!=-1&&pass[u][choose[v][0]]!=-1&&pass[u][choose[v][1]]!=-1){//两个都继承上来
                choose[u][0]=choose[v][0];
                choose[u][1]=choose[v][1];//直接采用儿子的限制
                return;
            }
        }
        int x=-1,y=-1;
        FOR(i,1,m)if(pass[u][i]!=-1)x==-1?x=i:y=i;
        choose[u][0]=x,choose[u][1]=y;
        //printf("solve(%d,%d)\n",u,p);
        //FOR(i,1,m)printf("%d%c",pass[u][i]," \n"[i==m]);
        //printf("x=%d,y=%d\n",x,y);
        g_lim[x].pb({y,pass[u][x]^pass[u][y]^1});
        g_lim[y].pb({x,pass[u][x]^pass[u][y]^1});
        //printf("add_lim(%d,%d,%d)\n",x,y,pass[u][x]^pass[u][y]^1);
    }
}
int rv[N],vis[N];
void dfs2(int u,int col){
    if(vis[u])return;
    rv[u]=col;
    vis[u]=1;
    for(pii p:g_lim[u])dfs2(p.fi,p.se^col);
}
int main(){
    scanf("%d%d",&n,&m);
    FOR(i,1,n-1){
        int a,b;
        scanf("%d%d",&a,&b);
        g[a].pb(b),g[b].pb(a);
    }
    dfs(1,0);
    FOR(i,1,m){
        scanf("%d %d",&u[i],&v[i]);
        w[i]=lca(u[i],v[i]);
        if(u[i]!=w[i])path[u[i]].pb({i,0});//0:从u出发的边
        if(v[i]!=w[i])path[v[i]].pb({i,1});//1:到达u的边
        path[w[i]].pb({i,-1});//经过u的边
    }
    solve(1,0);
    FOR(i,1,m)if(!vis[i])dfs2(i,0);
    printf("%d\n",ans);
    FOR(i,1,m){
        if(rv[i])printf("%d %d\n",v[i],u[i]);
        else printf("%d %d\n",u[i],v[i]);
    }
    return 0;
}