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, fa[2005], dep[2005], a[2005], b[2005], d[2005], cov[2005], st[2005], ed[2005], dfs_clock;
bool fix[2005];
vector<int> g[2005];

bool cover_dir(int i, int u) {
	return st[a[i]] >= st[u] && st[a[i]] <= ed[u] && (st[b[i]] < st[u] || st[b[i]] > ed[u]);
}
bool cover(int i, int u) {
	return (st[a[i]] >= st[u] && st[a[i]] <= ed[u]) ^ (st[b[i]] >= st[u] && st[b[i]] <= ed[u]);
}

void dfs(int u) {
	st[u] = ++dfs_clock;
	for (int v : g[u])
		if (v != fa[u])
			fa[v] = u, dep[v] = dep[u] + 1, dfs(v);
	ed[u] = dfs_clock;
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1, u, v; i < n; ++i) {
		scanf("%d%d", &u, &v);
		g[u].push_back(v), g[v].push_back(u);
	}
	dfs(1);
	for (int i = 1; i <= m; ++i) {
		scanf("%d%d", a + i, b + i);
		for (int u = a[i], v = b[i]; u != v;) dep[u] > dep[v] ? (++d[u], u = fa[u]) : (++d[v], v = fa[v]);
	}
	int ans = 0;
	for (int i = 2; i <= n; ++i) ans += min(2, d[i]);
	printf("%d\n", ans);
	for (int cnt = 1; cnt <= m; ++cnt) {
		int u = -1;
		for (int i = 2; i <= n; ++i) if (d[i] == 1 && (cov[i] == 1 || cov[i] == 2)) u = i;
		if (u == -1) for (int i = 2; i <= n; ++i) if (d[i] >= 2) u = i;
		for (int i = 1; i <= m; ++i)
			if (!fix[i] && cover(i, u)) {
				fix[i] = true;
				if (cov[u] == 2 ^ cover_dir(i, u)) swap(a[i], b[i]);
				for (int u = a[i], v = b[i]; u != v;)
					dep[u] > dep[v] ? (cov[u] |= 1, --d[u], u = fa[u]) : (cov[v] |= 2, --d[v], v = fa[v]);
				break;
			}
	}
	for (int i = 1; i <= m; ++i)
		printf("%d %d\n", a[i], b[i]);
	return 0;
}