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>

const int MAXN = 2010;
int head[MAXN], nxt[MAXN << 1], to[MAXN << 1], tot;
void addedge(int b, int e) {
	nxt[++tot] = head[b]; to[head[b] = tot] = e;
	nxt[++tot] = head[e]; to[head[e] = tot] = b;
}
int n, m, fa[MAXN], vis[MAXN];
std::vector<int> out[MAXN];
const int MAXP = MAXN * 5;
int ls[MAXP], rs[MAXP], bak, ansl[MAXP];
int decl[MAXP];
std::set<int> hav[MAXN];
int dep[MAXN];
void dfs(int u) {
	for (int i = head[u]; i; i = nxt[i])
		if (to[i] != fa[u]) {
			out[u].push_back(to[i]);
			dep[to[i]] = dep[u] + 1;
			fa[to[i]] = u;
			dfs(to[i]);
		}
}
int dc[MAXN];
int ans = 0;
int main() {
	std::cin >> n >> m; bak = m;
	for (int i = 1, t1, t2; i < n; ++i) {
		std::cin >> t1 >> t2;
		addedge(t1, t2);
	}
	dfs(1);
	for (int i = 1; i <= m; ++i) {
		std::cin >> ls[i] >> rs[i];
		hav[ls[i]].insert(i);
		hav[rs[i]].insert(i);
		int x = ls[i], y = rs[i];
		if (dep[x] < dep[y]) std::swap(x, y);
		while (dep[x] > dep[y]) ++dc[x], x = fa[x];
		while (x != y) ++dc[x], ++dc[y], x = fa[x], y = fa[y];
	}
	for (int i = 1; i <= n; ++i) ans += std::min(2, dc[i]);
	for (int i = 1; i < n; ++i) {
		int cho = 0;
		for (int j = 1; j <= n; ++j) if (!vis[j]) {
			int cx = 0;
			for (auto t : out[j]) cx += !vis[t];
			if (!cx) { cho = j; break; }
		}
		vis[cho] = true;
		std::vector<int> aq, dx;
		for (auto t : hav[cho]) {
			dx.push_back(t);
			if (ls[t] != rs[t]) aq.push_back(t);
		}
		for (auto t : dx) {
			hav[ls[t]].erase(t);
			hav[rs[t]].erase(t);
		}
	//	std::cout << "DBG " << cho << std::endl;
	//	for (auto t : aq) std::cout << '(' << ls[t] << ',' << rs[t] << ") ";
	//	std::cout << std::endl;
		while (aq.size() >= 2) {
			int tz = aq.size();
			int x = aq[tz - 1], y = aq[tz - 2];
			aq.erase(aq.end() - 2, aq.end());
			int now = ++bak;
			if (ls[x] == cho) {
				ls[now] = rs[x];
				decl[x] = -now;
			} else {
				ls[now] = ls[x];
				decl[x] = now;
			}
			if (rs[y] == cho) {
				rs[now] = ls[y];
				decl[y] = -now;
			} else {
				rs[now] = rs[y];
				decl[y] = now;
			}
			// std::cout << "MERGE " << ls[now] << ' ' << rs[now] << std::endl;
			hav[ls[now]].insert(now);
			hav[rs[now]].insert(now);
		}
		if (aq.size() == 0) continue;
		if (aq.size() == 1) {
			int t = aq[0], now = ++bak;
			ls[now] = ls[t], rs[now] = rs[t];
			decl[t] = now;
			if (ls[t] == cho) ls[now] = fa[cho];
			else if (rs[t] == cho) rs[now] = fa[cho];
			hav[ls[now]].insert(now);
			hav[rs[now]].insert(now);
		}
	}
	for (int i = bak; i; --i) if (decl[i])
		ansl[i] = decl[i] > 0 ? ansl[decl[i]] : 1 - ansl[-decl[i]];
	std::cout << ans << std::endl;
	for (int i = 1; i <= m; ++i) {
		if (ansl[i]) std::swap(ls[i], rs[i]);
		std::cout << ls[i] << ' ' << rs[i] << '\n';
	}
	return 0;
}