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:
/*
* @Author: wxyww
* @Date: 2020-03-31 11:02:37
* @Last Modified time: 2020-03-31 11:53:58
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<bitset>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
const int N = 4010;
ll read() {
	ll x=0,f=1;char c=getchar();
	while(c<'0'||c>'9') {
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}
struct node {
	int v,nxt;
}e[N];
int n,m,ejs,head[N],fa[N],dep[N],a[N],b[N],tot;
int du[N],c[N],f[N],flag[N],used[N];
void add(int u,int v) {
	e[++ejs].v = v;e[ejs].nxt = head[u];head[u] = ejs;
}
void dfs(int u,int father) {
	dep[u] = dep[father] + 1;
	fa[u] = father;
	for(int i = head[u];i;i = e[i].nxt) {
		int v = e[i].v;
		if(v == father) continue;
		dfs(v,u);
	}
}

void solve() {
	int u = 0;
	for(int i = 1;i <= n;++i) {
		if(flag[i]) continue;
		if(du[i] == 1) {
			u = i;break;
		}
	}


	if(!u) return;

	if(!c[u]) {
		for(int i = head[u];i;i = e[i].nxt) {
			int v = e[i].v;
			if(flag[v]) continue;
			du[v]--;
		}
		flag[u] = 1;
		solve();
		return;
	}

	if(c[u] == 1) {
		for(int i = head[u];i;i = e[i].nxt) {
			int v = e[i].v;
			if(flag[v]) continue;
			du[v]--;
		}
		flag[u] = 1;
		for(int i = 1;i <= tot;++i) {
			if(used[i]) continue;
			if(b[i] == u) swap(a[i],b[i]);
			if(a[i] != u) continue;
			int tmp = 0;
			for(int j = head[u];j;j = e[j].nxt) {
				int v = e[j].v;
				if(!flag[v]) {
					tmp = v; 
					break;
				}
			}
			if(b[i] != tmp) {
				a[i] = tmp;
				c[u]--;c[tmp]++;
				solve();
				if(a[i] == tmp) a[i] = u;
				else b[i] = u;
			}
			else {
				c[u]--;
				c[tmp]--;
				used[i] = 1;solve();
			}
			return;
		}
	}
	int x = 0,y = 0;
	for(int i = 1;i <= tot;++i) {
		if(used[i]) continue;
		if(b[i] == u) swap(a[i],b[i]);
		if(a[i] != u) continue;
		if(!x) x = i;
		else if(!y) y = i;
		else break;
	}
	used[x] = used[y] = 1;
	if(b[x] == b[y]) {
		swap(a[y],b[y]);
		c[u] -= 2;
		c[a[y]]--;c[b[x]]--;
		solve();
	}
	else {
		int now = ++tot;
		a[now] = b[x];
		b[now] = b[y];
		c[u] -= 2;
		solve();
		if(a[now] == b[x]) swap(a[x],b[x]);
		else swap(a[y],b[y]);
	}

}
int main() {
	// freopen("1.in","r",stdin);
	// freopen("me.out","w",stdout);
	n = read(),m = read();
	for(int i = 1;i < n;++i) {
		int u = read(),v = read();
		add(u,v);add(v,u);
		du[u]++;du[v]++;
	}
	dfs(1,0);

	for(int i = 1;i <= m;++i) {
		a[i] = read(),b[i] = read();
		if(dep[a[i]] < dep[b[i]]) swap(a[i],b[i]);
		int u = a[i],v = b[i];
		while(dep[u] > dep[v]) {
			f[u]++;u = fa[u];
		}
		while(u != v) {
			f[u]++;f[v]++;
			u = fa[u];v = fa[v];
		}
		c[a[i]]++;c[b[i]]++;
	}

	tot = m;
	solve();
	int ans = 0;
/*
	for(int i = 1;i <= n;++i) {
		printf("%d ",f[i]);
	}
	puts("");
*/

	for(int i = 1;i <= n;++i) ans += min(2,f[i]);
	printf("%d\n",ans);
	for(int i = 1;i <= m;++i)
		printf("%d %d\n",a[i],b[i]);
	return 0;
}