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:
//Created Time:2020年03月19日 星期四 09时02分09秒
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 5003

using namespace std;

int n, cnt, m;
int to[N << 3], s[N << 3], t[N << 3], deg[N], fa[N], dep[N], f[N], st[N];
bool vis[N << 3], book[N];

vector<int> e[N], w[N];

void dfs(int ,int);
void get(int ,int);

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i < n; ++i){
		int u, v; scanf("%d%d", &u, &v);
		e[u].push_back(v); e[v].push_back(u);
		++deg[u]; ++deg[v];
	} dfs(1, 0);
	for(int i = 1; i <= m; ++i){
		scanf("%d%d", s + i, t + i);
		to[i] = i; get(s[i], t[i]);
		w[s[i]].push_back(i);
		w[t[i]].push_back(i);
	}
	int ans = 0;
	for(int i = 2; i <= n; ++i)
		if(f[i] > 1) ans += 2;
		else if(f[i] == 1) ++ans;
	printf("%d\n", ans);
	queue<int> Q; int num = m;
	for(int i = 1; i <= n; ++i)
		if(deg[i] == 1) Q.push(i);
	while(!Q.empty()){
		int u = Q.front(), p; Q.pop(); book[u] = true;
		int sz = e[u].size();
		for(int i = 0; i < sz; ++i){
			int v = e[u][i]; --deg[v];
			if(deg[v] == 1) Q.push(v);
			if(!book[v]) p = v;
		}
		sz = w[u].size(); int top = 0;
		for(int i = 0; i < sz; ++i){
			int x = w[u][i]; if(vis[x]) continue;
			st[++top] = x; vis[x] = true;
		}
		if(top & 1){
			int x = st[top--];
			if(s[x] != u) swap(s[x], t[x]);
			if(p != t[x]){
				to[x] = ++num; to[num] = num;
				s[num] = p; t[num] = t[x];
				w[p].push_back(num);
				w[t[x]].push_back(num);
			}
		}
		for(int i = 1; i <= top; i += 2){
			int x = st[i], y = st[i + 1];
			if(s[x] != u) swap(s[x], t[x]);
			if(s[y] != u) swap(s[y], t[y]);
			if(t[x] == t[y]) swap(s[y], t[y]);
			else{
				to[x] = ++num; to[y] = num;
				s[num] = t[x]; t[num] = t[y];
				to[num] = num;
				w[t[x]].push_back(num);
				w[t[y]].push_back(num);
			}
		}
	}
	for(int i = num; i; --i){
		int x = to[i];
		if(s[x] == t[i]) swap(s[i], t[i]);
	}
	for(int i = 1; i <= m; ++i)
		printf("%d %d\n", s[i], t[i]);
	return 0;
}

void dfs(int u, int f){
	fa[u] = f; dep[u] = dep[f] + 1;
	int sz = e[u].size();
	for(int i = 0; i < sz; ++i){
		int v = e[u][i]; if(v == f) continue;
		dfs(v, u);
	} return ;
}

void get(int x, int y){
	if(dep[x] < dep[y]) swap(x, y);
	while(dep[x] > dep[y]) ++f[x], x = fa[x];
	if(x == y) return ;
	while(x != y) ++f[x], ++f[y], x = fa[x], y = fa[y];
	return ;
}