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 <iostream>
#include <vector>
#include <set>
#include <algorithm>

const int N = 4000 + 7;

int n, m;
int x[N], y[N];
std::set<int> s[N];
int z[N];
int p;

std::vector<int> e[N];
int f[N], d[N];
int dfn[N], dfc;
void dfs(int s) {
	std::sort(e[s].begin(), e[s].end());
	if (f[s]) e[s].erase(std::lower_bound(e[s].begin(), e[s].end(), f[s]));
	dfn[++dfc] = s;
	for (auto t : e[s]) f[t] = s, d[t] = d[s] + 1, dfs(t);
}

int vf[N];
int getvf(int x) {
	return vf[x] == x ? x : vf[x] = getvf(vf[x]);
}
void merge(int x, int y) {
	vf[getvf(x)] = getvf(y);
}

int exist(int x, int y) {
	return s[x].find(y * 2) != s[x].end() || s[x].find(y * 2 + 1) != s[x].end();
}

int main() {
//	freopen("code.in", "r", stdin);
//	freopen("code.out", "w", stdout);
	
	scanf("%d%d", &n, &m);
	for (int i = 1, u, v; i < n; ++i)
		scanf("%d%d", &u, &v), e[u].push_back(v), e[v].push_back(u);
	dfs(1);

	for (int i = 1; i <= m; ++i) {
		scanf("%d%d", x + i, y + i);
		for (int u = x[i], v = y[i]; u != v; )
			d[u] > d[v] ? (s[u].insert(i * 2), u = f[u]) : (s[v].insert(i * 2 + 1), v = f[v]);
	}
	for (int i = 1; i <= m * 2; ++i) vf[i] = i;

	for (int i = 1; i <= n; ++i)
		p += std::min(2, int(s[dfn[i]].size()));
	for (int i = n, j, k, l; i; --i) {
		if (s[dfn[i]].size() < 2 || z[dfn[i]]) continue;
		j = *s[dfn[i]].begin();
		k = *++s[dfn[i]].begin();
		l = (j ^ k) & 1 ^ 1;
		for (int q = 1; q <= n; ++q)
			if (exist(q, j / 2) && exist(q, k / 2))
				s[q].erase(j), s[q].erase(j ^ 1), s[q].erase(k), s[q].erase(k ^ 1), z[q] = 1;
		for (int q = 1; q <= n; ++q) {
			if (s[q].find(j) != s[q].end())
				s[q].erase(j), s[q].insert(k / 2 * 2 + (j & 1 ^ l));
			if (s[q].find(j ^ 1) != s[q].end())
				s[q].erase(j ^ 1), s[q].insert(k / 2 * 2 + (j & 1 ^ 1 ^ l));
		}
		merge(k / 2, j / 2 + m * l),
		merge(k / 2 + m, j / 2 + m * (l ^ 1));
	}

	printf("%d\n", p);
	for (int i = 1; i <= m; ++i) {
		if (getvf(1) != getvf(i) && getvf(1) != getvf(i + m)) merge(1, i);
		getvf(1) == getvf(i) ? printf("%d %d\n", x[i], y[i]) : printf("%d %d\n", y[i], x[i]);
	}
	return 0;
}