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 <iostream>
#include <cstring>
#define MN 2010

int cnt[MN], id[MN];
int fa[MN], dep[MN];
int u[MN], v[MN];
bool ok[MN], used[MN];
int h[MN], nxt[2 * MN], to[2 * MN], K = 0;

void ins(int u, int v) {nxt[++K] = h[u]; h[u] = K; to[K] = v;}

int lca(int u, int v)
{
	while(dep[u] > dep[v]) u = fa[u];
	while(dep[v] > dep[u]) v = fa[v];
	while(u != v) u = fa[u], v = fa[v];
	return u;
}

void modify(int o, int t)
{
	int l = lca(u[o], v[o]); 
	for(int i = u[o]; i != l; i = fa[i])
		cnt[i] += t, id[i] ^= o;
	for(int i = v[o]; i != l; i = fa[i])
		cnt[i] += t, id[i] ^= o;
	if(t == -1) used[o] = 1;
}

bool calc(int a, int b, int c, int d)
{
	if(std::max(dep[b], dep[d]) < dep[lca(a, c)]) return 1;
	else return 0;
}

void solve(int x, int y)
{
	int o = 0, l1 = lca(x, y);
	for(int i = x; i != l1; i = fa[i])
		if(cnt[i] == 1) o = id[i];
	for(int i = y; i != l1; i = fa[i])
		if(cnt[i] == 1) o = id[i];
	if(!o) return;
	int l2 = lca(u[o], v[o]);
	modify(o, -1); 
	if(calc(x, l1, u[o], l2)) ok[o] = 1, solve(x, u[o]), solve(v[o], y);
	else if(calc(x, l1, v[o], l2)) ok[o] = 0, solve(x, v[o]), solve(u[o], y);
	else if(calc(y, l1, u[o], l2)) ok[o] = 0, solve(u[o], y), solve(x, v[o]);
	else if(calc(y, l1, v[o], l2)) ok[o] = 1, solve(v[o], y), solve(x, u[o]);
}

void dfs(int x, int f)
{
	fa[x] = f; dep[x] = dep[f] + 1;
	for(int i = h[x]; i; i = nxt[i])
		if(to[i] != f) dfs(to[i], x);
}

int main()
{
	int n, m; scanf("%d%d", &n, &m);
	for(int i = 1; i < n; i++)
	{
		int a, b; scanf("%d%d", &a, &b);
		ins(a, b); ins(b, a);
	}
	dfs(1, 0);
	for(int i = 1; i <= m; i++) scanf("%d%d", &u[i], &v[i]), modify(i, 1);
	int ans = 0; for(int i = 2; i <= n; i++) ans += std::min(cnt[i], 2);
	for(int i = 1; i <= m; i++)
		if(!used[i]) modify(i, -1), solve(u[i], v[i]);
	printf("%d\n", ans);
	for(int i = 1; i <= m; i++)
	{
		if(ok[i]) printf("%d %d\n", v[i], u[i]);
		else printf("%d %d\n", u[i], v[i]);
	}
}