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<bits/stdc++.h>
using namespace std;
int n, m, svm, f[2007], dep[2007], v[2007], son[6007][2], ans[2007], res;
bool vis[6007];
pair<int, int>pp[6007];
vector<pair<int, int> >e[2007], ask[2007];
int read()
{
	int num = 0;
	char c = getchar();
	while (c < '0' || c>'9')c = getchar();
	while (c >= '0' && c <= '9')num = num * 10 + c - '0', c = getchar();
	return num;
}
int dfs(int w, int fa)
{
	f[w] = fa;
	dep[w] = dep[fa] + 1;
	for (int i = 0, lim = e[w].size(); i < lim; i++)
		if (e[w][i].first != fa)
			v[e[w][i].first] = e[w][i].second, dfs(e[w][i].first, w);
	return 0;
}
int glca(int a, int b)
{
	if (dep[a] < dep[b])swap(a, b);
	while (dep[a] > dep[b])a = f[a];
	while (a != b)a = f[a], b = f[b];
	return a;
}
int gline(int w)
{
	vis[w] = 1;
	if (son[w][0])
	{
		if (pp[son[w][0]].first != pp[w].first)
		{
			swap(pp[son[w][0]].first, pp[son[w][0]].second);
			swap(son[son[w][0]][0], son[son[w][0]][1]);
		}
		if (pp[son[w][0]].first != pp[w].first)cout << "ERROR" << endl;
		gline(son[w][0]);
	}
	if (son[w][1])
	{
		if (pp[son[w][1]].second != pp[w].second)
		{
			swap(pp[son[w][1]].first, pp[son[w][1]].second);
			swap(son[son[w][1]][0], son[son[w][1]][1]);
		}
		if (pp[son[w][1]].second != pp[w].second)cout << "ERROR" << endl;
		gline(son[w][1]);
	}
	return 0;
}
int main()
{
	n = read();
	m = svm = read();
	for (int i = 1; i < n; i++)
	{
		int a = read(), b = read();
		e[a].push_back(make_pair(b, i));
		e[b].push_back(make_pair(a, i));
	}
	for (int i = 1; i <= m; i++)
	{
		int a = pp[i].first = read(), b = pp[i].second = read();
		ask[a].push_back(make_pair(b, i));
		ask[b].push_back(make_pair(a, i));
	}
	for (int t = 2; t <= n; t++)
	{
		int p = 0;
		for (int i = 1; i <= n && !p; i++)
			if (e[i].size() == 1)
				p = i;
		dfs(p, 0);
		for (int lp = ask[p].size(); lp > 1; lp -= 2)
		{
			pair<int, int>pa = ask[p][lp - 1];
			pair<int, int>pb = ask[p][lp - 2];
			int a = pa.first, b = pb.first;
			int lca = glca(a, b);
			while (lca != p)ans[v[lca]] += 2, lca = f[lca];
			int nxt = ++m;
			pp[nxt].first = a, pp[nxt].second = b;
			son[nxt][0] = pa.second, son[nxt][1] = pb.second;
			*find(ask[a].begin(), ask[a].end(), make_pair(p, pa.second)) = make_pair(b, nxt);
			*find(ask[b].begin(), ask[b].end(), make_pair(p, pb.second)) = make_pair(a, nxt);
			ask[p].pop_back();
			ask[p].pop_back();
		}
		int fa = e[p][0].first;
		if (ask[p].size() == 1)
		{
			ans[e[p][0].second]++;
			int pos = ask[p][0].first, val = ask[p][0].second, nxt = ++m;
			son[nxt][0] = val;
			pp[nxt].first = pos, pp[nxt].second = fa;
			*find(ask[pos].begin(), ask[pos].end(), make_pair(p, val)) = make_pair(fa, nxt);
			ask[fa].push_back(make_pair(pos, nxt));
		}
		e[fa].erase(find(e[fa].begin(), e[fa].end(), make_pair(p, e[p][0].second)));
		e[p].clear();
		ask[p].clear();
	}
	for (int i = m; i >= 1; i--)
		if (!vis[i])
			gline(i);
	for (int i = 1; i < n; i++)
		res += min(ans[i], 2);
	cout << res << endl;
	for (int i = 1; i <= svm; i++)
		printf("%d %d\n", pp[i].first, pp[i].second);
	return 0;
}