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 maxn = 2000;
int n, m, fa[maxn + 3], dep[maxn + 3], u[maxn + 3], v[maxn + 3], w[maxn + 3], ans;
int b[maxn + 3][maxn + 3], g[maxn + 3][2], tot, lft[maxn + 3], rht[maxn + 3], wei[maxn + 3];
bool vis[maxn + 3], res[maxn + 3];
vector<int> G[maxn + 3], D[maxn + 3], V[maxn + 3];
vector<pair<int, int> > S[maxn + 3];

void add(int u, int v) {
	G[u].push_back(v);
}

void dfs(int u, int pa = 0) {
	for (int i = 0, v; i < G[u].size(); i++) {
		v = G[u][i];
		if (v == pa) continue;
		dep[v] = dep[u] + 1;
		fa[v] = u;
		dfs(v, u);
	}
}

void add_lim(int u, int v, int w) {
	lft[++tot] = u, rht[tot] = v, wei[tot] = w;
}

void solve(int u, int pa = 0) {
	g[u][0] = g[u][1] = -1;
	for (int i = 1; i <= m; i++) {
		b[u][i] = -1;
	}
	for (int i = 0, v; i < G[u].size(); i++) {
		v = G[u][i];
		if (v == pa) continue;
		solve(v, u);
		for (int j = 1; j <= m; j++) {
			if (b[u][j] == -1) {
				b[u][j] = b[v][j];
			}
		}
	}
	for (int i = 0; i < S[u].size(); i++) {
		pair<int, int> x = S[u][i];
		b[u][x.first] = x.second;
	}
	int cnt = 0;
	for (int i = 1; i <= m; i++) {
		cnt += b[u][i] != -1;
	}
	if (cnt < 2) {
		ans += cnt;
	} else {
		ans += 2;
		int p = 0;
		for (int i = 0, v; i < G[u].size(); i++) {
			v = G[u][i];
			if (v == pa) continue;
			if (g[v][0] == -1) continue;
			if (~b[u][g[v][0]] && ~b[u][g[v][1]]) {
				p = v;
				break;
			}
		}
		if (p) {
			g[u][0] = g[p][0];
			g[u][1] = g[p][1];
		} else {
			int x = 0, y = 0;
			for (int i = 1; i <= m; i++) {
				if (b[u][i] == -1) continue;
				if (!x) {
					x = i;
				} else if (x && !y) {
					y = i;
				}
			}
			g[u][0] = x, g[u][1] = y;
			add_lim(x, y, b[u][x] ^ b[u][y] ^ 1);
		}
	}
}

void color(int u) {
	vis[u] = true;
	for (int i = 0, v; i < D[u].size(); i++) {
		v = D[u][i];
		if (vis[v]) continue;
		res[v] = res[u] ^ V[u][i];
		color(v);
	}
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1, u, v; i < n; i++) {
		scanf("%d %d", &u, &v);
		add(u, v), add(v, u);
	}
	dfs(1);
	for (int i = 1; i <= m; i++) {
		scanf("%d %d", &u[i], &v[i]);
		int x = u[i], y = v[i];
		while (x != y) {
			if (dep[x] < dep[y]) swap(x, y);
			x = fa[x];
		}
		w[i] = x;
		if (u[i] != w[i]) {
			S[u[i]].push_back(make_pair(i, 0));
		}
		if (v[i] != w[i]) {
			S[v[i]].push_back(make_pair(i, 1));
		}
		S[w[i]].push_back(make_pair(i, -1));
	}
	solve(1);
	printf("%d\n", ans);
	for (int i = 1; i <= tot; i++) {
		D[lft[i]].push_back(rht[i]), V[lft[i]].push_back(wei[i]);
		D[rht[i]].push_back(lft[i]), V[rht[i]].push_back(wei[i]);
	}
	for (int i = 1; i <= m; i++) {
		if (!vis[i]) color(i);
	}
	for (int i = 1; i <= m; i++) {
		if (!res[i]) {
			printf("%d %d\n", u[i], v[i]);
		} else {
			printf("%d %d\n", v[i], u[i]);
		}
	}
	return 0;
}