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<algorithm>
#include<cstring>
using namespace std;
struct Edge {
	int to;
	int nxt;
} e[4005];
int n, m, edgenum, head[2005], pa[2005], dep[2005], a[4005], b[4005], tot;
int d[2005], c[2005], f[2005];
bool flag[2005], used[4005];
void add(int u, int v) {
	e[++edgenum].to = v;
	e[edgenum].nxt = head[u];
	head[u] = edgenum;
}
void dfs(int node) {
	dep[node] = dep[pa[node]] + 1;
	for (int hd = head[node]; hd; hd = e[hd].nxt) {
		int to = e[hd].to;
		if (to == pa[node])continue;
		pa[to] = node;
		dfs(to);
	}
}
void solve() {
	int u = 0;
	for (int i = 1; i <= n; i++) {
		if (flag[i])continue;
		if (d[i] == 1) {
			u = i;
			break;
		}
	}
	if (u == 0)return;
	if (c[u] == 0) {
		for (int hd = head[u]; hd; hd = e[hd].nxt) {
			int to = e[hd].to;
			if (flag[to])continue;
			d[to]--;
		}
		flag[u] = 1;
		solve();
		return;
	}
	if (c[u] == 1) {
		for (int hd = head[u]; hd; hd = e[hd].nxt) {
			int to = e[hd].to;
			if (flag[to])continue;
			d[to]--;
		}
		flag[u] = 1;
		for (int i = 1; i <= tot; i++) {
			if (used[i])continue;
			if (b[i] == u)swap(a[i], b[i]);
			if (a[i] != u)continue;
			int v = 0;
			for (int hd = head[u]; hd; hd = e[hd].nxt) {
				int to = e[hd].to;
				if (!flag[to]) {
					v = to;
					break;
				}
			}
			if (b[i] != v) {
				a[i] = v;
				c[u]--, c[v]++;
				solve();
				if (a[i] == v)a[i] = u;
				else b[i] = u;
			} else {
				c[u]--, c[v]--;
				used[i] = 1;
				solve();
			}
			return;
		}
	}
	int x = 0, y = 0;
	for (int i = 1; i <= tot; i++) {
		if (used[i])continue;
		if (b[i] == u)swap(a[i], b[i]);
		if (a[i] != u)continue;
		if (!x)x = i;
		else if (!y)y = i;
		else break;
	}
	used[x] = used[y] = 1;
	if (b[x] == b[y]) {
		swap(a[y], b[y]);
		c[u] -= 2;
		c[a[y]]--, c[b[x]]--;
		solve();
	} else {
		int now = ++tot;
		a[now] = b[x];
		b[now] = b[y];
		c[u] -= 2;
		solve();
		if (a[now] == b[x])swap(a[x], b[x]);
		else swap(a[y], b[y]);
	}
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i < n; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		add(u, v);
		add(v, u);
		d[u]++, d[v]++;
	}
	dfs(1);
	for (int i = 1; i <= m; i++) {
		scanf("%d%d", &a[i], &b[i]);
		if (dep[a[i]] < dep[b[i]])swap(a[i], b[i]);
		int u = a[i], v = b[i];
		while (dep[u] > dep[v]) {
			f[u]++;
			u = pa[u];
		}
		while (u != v) {
			f[u]++, f[v]++;
			u = pa[u], v = pa[v];
		}
		c[a[i]]++, c[b[i]]++;
	}
	tot = m;
	solve();
	tot = 0;
	for (int i = 1; i <= n; i++)tot += min(2, f[i]);
	printf("%d\n", tot);
	for (int i = 1; i <= m; i++)printf("%d %d\n", a[i], b[i]);
	return 0;
}