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
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
template<typename T> inline void chkmin(T &a, const T &b) { a = a < b ? a : b; }
template<typename T> inline void chkmax(T &a, const T &b) { a = a > b ? a : b; }

const int MAXN = 2005, MAXM = 10005;
int visp[MAXN], deg[MAXN], uu[MAXN], vv[MAXN], vise[MAXM], arr[MAXN];
int par[MAXN], que[MAXN], fr[MAXM], to[MAXM], cnt[MAXN], n, m, tot;
int nxt[MAXM], dir[MAXM], dep[MAXM], dfn[MAXN], lg[MAXM];
vector<int> gra[MAXN], vec[MAXN];

int tmp[MAXN][MAXN];

void dfs1(int u, int fa) {
	par[u] = fa, dep[u] = dep[fa] + 1;
	for (int v : gra[u]) if (v != fa) dfs1(v, u);
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i < n; i++) {
		int u, v; scanf("%d%d", &u, &v);
		gra[u].push_back(v), gra[v].push_back(u);
		++deg[u], ++deg[v];
	}
	dfs1(1, 0);
	for (int i = 1; i <= m; i++) {
		int u, v; scanf("%d%d", &u, &v);
		uu[i] = fr[i] = u, vv[i] = to[i] = v;
		vec[u].push_back(i), vec[v].push_back(i);
		if (dep[u] > dep[v]) swap(u, v);
		while (dep[v] > dep[u]) ++cnt[v], v = par[v];
		while (v != u) ++cnt[v], ++cnt[u], v = par[v], u = par[u];
	}
	int ans = 0, all = m;
	for (int i = 1; i <= n; i++) ans += min(cnt[i], 2);
	int he = 1, ta = 0;
	for (int i = 1; i <= n; i++) if (deg[i] == 1) que[++ta] = i;
	
	while (he <= ta) {
		int u = que[he++];
		visp[u] = 1;
		for (int v : gra[u]) if (!visp[v]) {
			if (--deg[v] == 1) que[++ta] = v;
			int cnt = 0;
			for (int i : vec[u])
				if (!vise[i] && fr[i] != to[i]) arr[++cnt] = i;
			for (int i = 1; i < cnt; i += 2) {
				int a = arr[i], b = arr[i + 1];
				vise[a] = vise[b] = 1, ++all;
				int p = fr[a] == u ? to[a] : fr[a];
				int q = fr[b] == u ? to[b] : fr[b];
				fr[all] = p, to[all] = q;
				nxt[a] = nxt[b] = all;
				dir[a] = fr[a] == u, dir[b] = to[b] == u;
				vec[p].push_back(all);
				vec[q].push_back(all);
			}
			if(cnt & 1) {
				int p = arr[cnt];
				(fr[p] == u ? fr[p] : to[p]) = v;
				vec[v].push_back(p);
			}
		}
		vec[u].clear();
	}
	for (int i = all; i > 0; i--) {
		if (nxt[i]) dir[i] ^= dir[nxt[i]];
		if (i <= m && dir[i]) swap(uu[i], vv[i]);
	}
	printf("%d\n", ans);
	for (int i = 1; i <= m; i++) printf("%d %d\n", uu[i], vv[i]);
	return 0;
}