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 meow(args...) fprintf(stderr, args)
typedef unsigned u32;
typedef long long s64;
typedef unsigned long long u64;
template<class T1, class T2> inline bool cmin(T1 &a, const T2 &b) {return b<a?(a=b, true):false;}
template<class T1, class T2> inline bool cmax(T1 &a, const T2 &b) {return a<b?(a=b, true):false;}
template<class Type> Type read() {
	Type a;
	bool b;
	unsigned char c;
	while(c=getchar()-48, (c>9)&(c!=253));
	for(a=(b=c==253)?0:c; (c=getchar()-48)<=9; a=a*10+c);
	return b?-a:a;
}
int (*rd)()=read<int>;

const int N=1e5+1;
int n, m, u[N], v[N], u0[N], v0[N], follow[N], swap[N], ans, dep[N], euler_n, el[N], st[12][2*N], *euler=st[0], cnt[N], deg[N], queue[N];
bool vis[N];
std::vector<int> edge[N], path[N];
void dfs(int u, int p, int d) {
	euler[el[u]=euler_n++]=u;
	dep[u]=d;
	for(int v: edge[u])
		if(v!=p) {
			dfs(v, u, d+1);
			euler[euler_n++]=u;
		}
}
int lca(int x, int y) {
	x=el[x], y=el[y];
	if(x>y) std::swap(x, y);
	int lv=31-__builtin_clz(y-x+1);
	x=st[lv][x], y=st[lv][y-(1<<lv)+1];
	return dep[x]<dep[y]?x:y;
}
void dfs2(int u, int p) {
	for(int v: edge[u])
		if(v!=p) {
			dfs2(v, u);
			cnt[u]+=cnt[v];
		}
	ans+=std::min(cnt[u], 2);
}
int find(int x) {
	if(follow[x]==0) return x;
	int y=find(follow[x]);
	swap[x]^=swap[follow[x]];
	return follow[x]=y;
}
int main() {
	n=rd(), m=rd();
	for(int i=1; i<n; ++i) {
		int a=rd(), b=rd();
		edge[a].push_back(b);
		edge[b].push_back(a);
		++deg[a], ++deg[b];
	}
	dfs(1, 0, 0);
	for(int i=1; (1<<i)<=euler_n; ++i)
		for(int j=0; (1<<i)+j<=euler_n; ++j) {
			int x=st[i-1][j], y=st[i-1][(1<<(i-1))+j];
			st[i][j]=dep[x]<dep[y]?x:y;
		}
	for(int i=1; i<=m; ++i) {
		u[i]=u0[i]=rd();
		v[i]=v0[i]=rd();
		path[u[i]].push_back(i);
		path[v[i]].push_back(i);
		cnt[lca(u[i], v[i])]-=2;
		++cnt[u[i]], ++cnt[v[i]];
	}
	dfs2(1, 0);
	int *qhead=queue, *qtail=queue;
	for(int i=1; i<=n; ++i) if(deg[i]==1) *qtail++=i;
	for(int i=1; i<n; ++i) {
		int x=*qhead++, y;
		std::vector<int> t;
		vis[x]=true;
		for(int j: edge[x]) if(!vis[j]) y=j;
		for(int j: path[x])
			if(u[j]!=v[j]&&follow[j]==0)
				t.push_back(j);
		for(int j=0; j+1<t.size(); j+=2) {
			int p=t[j], q=t[j+1];
			follow[q]=p;
			swap[q]=(u[p]==x)^(u[q]==x)^1;
			if(u[p]==x) {
				u[p]^=u[q]^v[q];
				path[u[p]].push_back(p);
			} else {
				v[p]^=u[q]^v[q];
				path[v[p]].push_back(p);
			}
		}
		if(t.size()%2==1) {
			int p=t.back();
			if(u[p]==x) u[p]=y; else v[p]=y;
			path[y].push_back(p);
		}
		if(--deg[y]==1) *qtail++=y;
	}
	printf("%d\n", ans);
	for(int i=1; i<=m; ++i) {
		find(i);
		if(swap[i]) std::swap(u0[i], v0[i]);
		printf("%d %d\n", u0[i], v0[i]);
	}
	return 0;
}