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>

template <class T>
inline void read(T &res)
{
	res = 0; bool bo = 0; char c;
	while (((c = getchar()) < '0' || c > '9') && c != '-');
	if (c == '-') bo = 1; else res = c - 48;
	while ((c = getchar()) >= '0' && c <= '9')
		res = (res << 3) + (res << 1) + (c - 48);
	if (bo) res = ~res + 1;
}

typedef std::vector<int> vi;
typedef std::vector<std::pair<int, int> > vpi;

const int N = 2005, M = 4005;

int n, m, ecnt, nxt[M], adj[N], go[M], fa[N], dep[N], d[N], pt[N], a[N], ans;

void add_edge(int u, int v)
{
	nxt[++ecnt] = adj[u]; adj[u] = ecnt; go[ecnt] = v;
	nxt[++ecnt] = adj[v]; adj[v] = ecnt; go[ecnt] = u;
}

void dfs(int u, int fu)
{
	dep[u] = dep[fa[u] = fu] + 1;
	for (int e = adj[u], v; e; e = nxt[e])
		if ((v = go[e]) != fu) dfs(v, u);
}

void getmark(int u, int v)
{
	while (dep[u] != dep[v])
		if (dep[u] > dep[v]) a[u]++, u = fa[u];
		else a[v]++, v = fa[v];
	while (u != v) a[u]++, a[v]++, u = fa[u], v = fa[v];
}

vi jiejuediao(vi V, vpi E, vpi path)
{
	vi res;
	for (int i = 0; i < path.size(); i++) res.push_back(0);
	if (V.size() == 1) return res;
	vi nV; vpi nE, nP; memset(d, 0, sizeof(d));
	int p1 = -1, p2 = -1;
	for (int i = 0; i < E.size(); i++) d[E[i].first]++, d[E[i].second]++,
		pt[E[i].first] = pt[E[i].second] = i;
	int u, v, e;
	for (int i = 0; i < V.size(); i++) if (d[V[i]] == 1)
	{
		u = V[i]; e = pt[u];
		v = E[e].first == u ? E[e].second : E[e].first;
		break;
	}
	for (int i = 0; i < path.size(); i++)
		if (path[i].first == u && path[i].second == u)
		{
			for (int j = 0; j < path.size(); j++)
				if (j != i) nP.push_back(path[j]);
			res = jiejuediao(V, E, nP); res.push_back(0);
			for (int j = path.size() - 1; j > i; j--)
				res[j] = res[j - 1], res[j - 1] = 0;
			return res;
		}
	for (int i = 0; i < path.size(); i++)
		if (path[i].first == u || path[i].second == u)
			(p1 == -1 ? p1 : p2) = i;
	if (p1 == -1)
	{
		for (int i = 0; i < V.size(); i++) if (u != V[i]) nV.push_back(V[i]);
		for (int i = 0; i < E.size(); i++) if (e != i) nE.push_back(E[i]);
		return jiejuediao(nV, nE, path);
	}
	if (p2 == -1)
	{
		for (int i = 0; i < path.size(); i++)
		{
			if (path[i].first == u) path[i].first = v;
			if (path[i].second == u) path[i].second = v;
		}
		return jiejuediao(V, E, path);
	}
	for (int i = 0; i < path.size(); i++) if (i != p1 && i != p2)
		nP.push_back(path[i]);
	nP.push_back(std::make_pair(path[p1].first == u ? path[p1].second : path[p1].first,
		path[p2].first == u ? path[p2].second : path[p2].first));
	vi tmp = jiejuediao(V, E, nP);
	for (int i = 0; i < path.size(); i++)
		if (i != p1 && i != p2) res[i] = tmp[i - (i > p1) - (i > p2)];
		else if (i == p1) res[i] = tmp[tmp.size() - 1] ^ (path[i].first == u);
		else res[i] = tmp[tmp.size() - 1] ^ (path[i].second == u);
	return res;
}

int main()
{
	vi V; vpi E, path;
	int x, y;
	read(n); read(m);
	for (int i = 1; i <= n; i++) V.push_back(i);
	for (int i = 1; i < n; i++) read(x), read(y), add_edge(x, y),
		E.push_back(std::make_pair(x, y));
	dfs(1, 0);
	for (int i = 1; i <= m; i++) read(x), read(y),
		path.push_back(std::make_pair(x, y)), getmark(x, y);
	for (int i = 2; i <= n; i++) ans += a[i] > 2 ? 2 : a[i];
	std::cout << ans << std::endl;
	vi res = jiejuediao(V, E, path);
	for (int i = 0; i < m; i++)
		if (res[i]) printf("%d %d\n", path[i].second, path[i].first);
		else printf("%d %d\n", path[i].first, path[i].second);
	return 0;
}