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
#include <iostream>
#include <vector>
#include <cstdio>
#define maxn 1000006
#define mp make_pair
#define fi first
#define se second
using namespace std;
vector <int> G[maxn];
vector < pair <int, int> > V[maxn];
pair <int, int> edges[maxn];
bool del[maxn];
int to[maxn], turn[maxn], f[maxn];
void addpath(int u, int v, int i){
	V[u].push_back(mp(v, i));
	V[v].push_back(mp(u, i));
	edges[i] = mp(u, v);
}
int dfn[maxn], dfn_cnt = 0, fa[maxn], ans;
void dfs(int u, int f){
	fa[u] = f;
	for (int i = 0; i < G[u].size(); ++ i){
		int v = G[u][i];
		if (v != f)
			dfs(v, u);
	}
	dfn[++ dfn_cnt] = u;
}
bool vis[2003][2003];
bool cover(int u, int e, int f){
	if (u == e)
		return true;
	for (int i = 0; i < G[u].size(); ++ i){
		int v = G[u][i];
		if (v != f && cover(v, e, u)){
			if (! vis[u][v]){
				++ ans;
				vis[u][v] = true;
			}
			return true;
		}
	}
	return false;
}
int read(){
	int cnt = 0;
	char ch = getchar();
	while (ch < '0' || '9' < ch)
		ch = getchar();
	while ('0' <= ch && ch <= '9'){
		cnt = cnt * 10 + ch - '0';
		ch = getchar();
	}
	return cnt;
}
int main(){
	int n = read(), m = read();
	int tmp = m;
	for (int i = 1; i < n; ++ i){
		int u = read(), v = read();
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for (int i = 1; i <= m; ++ i){
		int u = read(), v = read();
		addpath(u, v, i);
	}
	dfs(1, 0);
	for (int i = 1; i <= n; ++ i){
		int u = dfn[i], last = -1;
		del[u] = true;
		for (int j = 0; j < V[u].size(); ++ j){
			int v = V[u][j].fi;
			if (del[v])
				continue;
			if (last == -1){
				last = j;
				continue;
			}
			addpath(V[u][last].fi, V[u][j].fi, ++ m);
			to[V[u][j].se] = m;
			turn[V[u][j].se] = mp(u, v) != edges[V[u][j].se];
			to[V[u][last].se] = m;
			turn[V[u][last].se] = (mp(u, V[u][last].fi) != edges[V[u][last].se]) ^ 1;
			last = -1;
		}
		if (last != -1){
			addpath(fa[u], V[u][last].fi, ++ m);
			to[V[u][last].se] = m;
			turn[V[u][last].se]= mp(u, V[u][last].fi) != edges[V[u][last].se];
		}
	}
	for (int i = m; i; -- i){
		if (to[i])
			f[i] = f[to[i]] ^ turn[i];
		if (f[i])
			swap(edges[i].fi, edges[i].se);
	}
	for (int i = 1; i <= tmp; ++ i)
		cover(edges[i].fi, edges[i].se, 0);
	cout << ans << endl;
	for (int i = 1; i <= tmp; ++ i)
		printf("%d %d\n", edges[i].fi, edges[i].se);
}
