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;

const int N = 2005;

int n, m;
int res;
int sum[N];
int fr[N], to[N];
int h[N], p[11][N];
int color[N * 2];
vector<int> G[N], G2[N * 2];
set< pair<int, int> > S[N];
vector< pair<int, int> > vdel[N];

int lca(int u, int v) {
	if (h[u] > h[v]) swap(u, v);
	for (int i = 10; i >= 0; --i) if (h[p[i][v]] >= h[u]) v = p[i][v];
	for (int i = 10; i >= 0; --i) if (p[i][v] != p[i][u]) v = p[i][v], u = p[i][u];
	return (u == v) ? u : p[0][u];
}

void dfs(int u) {
	for (int i = 1; i <= 10; ++i) {
		p[i][u] = p[i - 1][p[i - 1][u]];
	}
	for (auto v : G[u]) if (p[0][u] != v) {
		p[0][v] = u, h[v] = h[u] + 1, dfs(v);
	}
}

void solve(int u) {
	int pmx = 0;
	for (auto v : G[u]) if (p[0][u] != v) {
		solve(v);
		if (S[v].size() == 1) res++;
		if (S[v].size() > 1) {
			res += 2;
			if (!sum[v]) {
				auto p0 = S[v].begin();
				auto p1 = ++S[v].begin();
				G2[p0 -> second].push_back(p1 -> second);
				G2[p1 -> second].push_back(p0 -> second);
				int x = p0 -> first;
				int y = p1 -> first;
				if (h[x] > h[y]) swap(x, y);
				sum[v]++, sum[y]--;
			}
		}
		sum[u] += sum[v];
		if (!pmx || S[pmx].size() < S[v].size()) pmx = v;
	} 
	if (pmx) swap(S[u], S[pmx]);
	for (auto v : G[u]) if (p[0][u] != v) {
		for (auto i : S[v]) S[u].insert(i);
	}
	for (auto i : vdel[u]) S[u].erase(i);
}

void dfscolor(int u) {
	for (auto v : G2[u]) {
		if (color[v]) {
			assert(color[v] == 3 ^ color[u]);
			continue;
		}
		color[v] = 3 ^ color[u], dfscolor(v);
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> m;
	for (int i = 1; i < n; ++i) {
		int u, v; cin >> u >> v;
		G[u].push_back(v), G[v].push_back(u);
	}
	p[0][1] = 1, dfs(1);
	for (int i = 0; i < m; ++i) {
		cin >> fr[i] >> to[i];
		if (h[fr[i]] > h[to[i]]) swap(fr[i], to[i]);
		int tmp = lca(fr[i], to[i]);
		if (tmp == fr[i]) {
			S[to[i]].insert({tmp, i * 2});
			vdel[tmp].push_back({tmp, i * 2});
		}
		else {
			S[to[i]].insert({tmp, i * 2});
			S[fr[i]].insert({tmp, i * 2 + 1});
			G2[i * 2].push_back(i * 2 + 1);
			G2[i * 2 + 1].push_back(i * 2);
			vdel[tmp].push_back({tmp, i * 2});
			vdel[tmp].push_back({tmp, i * 2 + 1});
		}
	}
	solve(1);
	for (int i = 0; i < m * 2; ++i) {
		if (!color[i]) color[i] = 1, dfscolor(i);
	}
	cout << res << '\n';
	for (int i = 0; i < m; ++i) {
		if (color[i * 2] == 1) cout << fr[i] << ' ' << to[i] << '\n';
		else cout << to[i] << ' ' << fr[i] << '\n';
	}
}