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 <cstring>
#include <cstdio>
#include <vector>
#define pb push_back
int lst[2005], to[4005], pre[4005], tot;
inline void add_edge(int u, int v)
{
	to[tot] = v;
	pre[tot] = lst[u];
	lst[u] = tot++;
}
template <typename T>
bool dfs(int u, int en, T &seq, int fa = -1)
{
	if (u == en)
		return true;
	for (int i = lst[u]; ~i; i = pre[i])
	{
		if (to[i] == fa)
			continue;
		seq.pb(i);
		if (dfs(to[i], en, seq, u))
			return true;
		seq.pop_back();
	}
	return false;
}
std::vector<int> path[2005], pass[2005];
int xp[2005], yp[2005], cnt[2005];
bool vis[4005], fix[2005];
int main()
{
	// freopen("AGC025-E.in", "r", stdin);
	memset(lst, -1, sizeof(lst));
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 1; i < n; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		add_edge(--u, --v);
		add_edge(v, u);
	}
	for (int i = 0; i < m; i++)
	{
		scanf("%d%d", xp + i, yp + i);
		dfs(--xp[i], --yp[i], path[i]);
		for (int e : path[i])
		{
			cnt[e >> 1]++;
			pass[e >> 1].pb(i);
		}
	}
	int ans = 0;
	for (int i = 0; i < tot >> 1; i++)
		ans += std::min(2, cnt[i]);
	printf("%d\n", ans);
	for (int t = 0; t < m; t++)
	{
		int idx = -1;
		for (int i = 0; i < tot >> 1 && -1 == idx; i++)
		{
			if (cnt[i] == 1 && vis[i << 1] != vis[i << 1 | 1])
				idx = vis[i << 1] ? i << 1 | 1 : i << 1;
		}
		for (int i = 0; i < tot >> 1 && -1 == idx; i++)
		{
			if (cnt[i])
				idx = i << 1;
		}
		int e = idx >> 1;
		while (fix[pass[e].back()])
			pass[e].pop_back();
		int x = pass[e].back();
		fix[x] = true;
		bool rev;
		for (int i : path[x])
		{
			if (i >> 1 == e)
				rev = i != idx;
		}
		if (rev)
		{
			std::swap(xp[x], yp[x]);
			for (int &i : path[x])
				i ^= 1;
		}
		for (int i : path[x])
		{
			vis[i] = true;
			cnt[i >> 1]--;
		}
	}
	for (int i = 0; i < m; i++)
		printf("%d %d\n", xp[i] + 1, yp[i] + 1);
	return 0;
}