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>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MOD = 998244353;

int n, m;
vector<int> G[2005];
int par[2005], dep[2005];
int eu[2005], ev[2005];
int eid[2005][2005];
bool hv[2005][2005];
int qu[2005], qv[2005];
int cnt[2005], xv[2005];

void dfs(int v, int p, int d)
{
	par[v] = p;
	dep[v] = d;
	rep(i, G[v].size()) {
		int u = G[v][i];
		if(u == p) continue;
		dfs(u, v, d + 1);
	}
}

bool vis[2005], dir[2005];
void dfs2(int cur)
{
	vis[cur] = true;
	int u = qu[cur], v = qv[cur];
	while(u != v) {
		if(dep[u] > dep[v]) {
			cnt[eid[u][par[u]]] --;
			xv[eid[u][par[u]]] ^= eu[eid[u][par[u]]] == u ? cur : ~cur;
			if(dir[cur]) hv[par[u]][u] = true; else hv[u][par[u]] = true;
			u = par[u];
		} else {
			cnt[eid[v][par[v]]] --;
			xv[eid[v][par[v]]] ^= eu[eid[v][par[v]]] == v ? ~cur : cur;
			if(dir[cur]) hv[v][par[v]] = true; else hv[par[v]][v] = true;
			v = par[v];
		}
	}
	u = qu[cur]; v = qv[cur];
	while(u != v) {
		if(dep[u] > dep[v]) {
			if(cnt[eid[u][par[u]]] == 1) {
				int nc = xv[eid[u][par[u]]];
				if(nc < 0) nc = ~nc;
				if(!hv[u][par[u]] || !hv[par[u]][u]) {
					dir[nc] = dir[cur] ^ 1 ^ (xv[eid[u][par[u]]] < 0) ^ eu[eid[u][par[u]]] != u;
					dfs2(nc);
				}
			}
			u = par[u];
		} else {
			if(cnt[eid[v][par[v]]] == 1) {
				int nc = xv[eid[v][par[v]]];
				if(nc < 0) nc = ~nc;
				if(!hv[v][par[v]] || !hv[par[v]][v]) {
					dir[nc] = dir[cur] ^ 1 ^ (xv[eid[v][par[v]]] < 0) ^ eu[eid[v][par[v]]] == v;
					dfs2(nc);
				}
			}
			v = par[v];
		}
	}
}

int main()
{
	scanf("%d%d", &n, &m);
	rep(i, n - 1) {
		scanf("%d%d", &eu[i], &ev[i]);
		G[eu[i]].push_back(ev[i]);
		G[ev[i]].push_back(eu[i]);
		eid[eu[i]][ev[i]] = eid[ev[i]][eu[i]] = i;
	}
	rep(i, m) scanf("%d%d", &qu[i], &qv[i]);
	dfs(1, -1, 0);
	
	rep(i, m) {
		int u = qu[i], v = qv[i];
		while(u != v) {
			if(dep[u] > dep[v]) {
				cnt[eid[u][par[u]]] ++;
				xv[eid[u][par[u]]] ^= eu[eid[u][par[u]]] == u ? i : ~i;
				u = par[u];
			} else {
				cnt[eid[v][par[v]]] ++;
				xv[eid[v][par[v]]] ^= eu[eid[v][par[v]]] == v ? ~i : i;
				v = par[v];
			}
		}
	}
	
	int ans = 0;
	rep(i, n - 1) ans += min(cnt[i], 2);
	printf("%d\n", ans);
	rep(i, m) if(!vis[i]) dfs2(i);
	rep(i, m) {
		if(dir[i]) swap(qu[i], qv[i]);
		printf("%d %d\n", qu[i], qv[i]);
	}
	return 0;
}