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 <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <vector>
inline int getint()
{
	int r = 0,s = 0;char c = getchar();for(;!isdigit(c);c = getchar()) s |= c == '-';
	for(;isdigit(c);c = getchar()) r = (((r << 2) + r) << 1) + (c ^ '0');return s ? -r : r;
}
const int N = 2020;
int head[N],next[N << 1],to[N << 1],tot = 1;
inline void addedge(int u,int v)
{
	next[++tot] = head[u];to[tot] = v;head[u] = tot;
	next[++tot] = head[v];to[tot] = u;head[v] = tot;
}
int S[N],T[N],cnt[N];
std::vector<int> path[N],edge[N];
int n,m;
inline int dfs(int u,int f,int de,int id)
{
	if(u == de) return 1;
	for(int i = head[u],v = to[i];~i;v = to[i = next[i]])
	{
		if(v == f) continue;
		++cnt[i >> 1];path[id].push_back(i);edge[i >> 1].push_back(id);
		if(dfs(v,u,de,id)) return 1;
		--cnt[i >> 1];path[id].pop_back();edge[i >> 1].pop_back();
	}
	return 0;
}
int vis[N << 1],used[N];
int main()
{
	memset(head,-1,sizeof(head));
	n = getint(),m = getint();
	for(int i = 1;i < n;i++) addedge(getint(),getint());
	for(int i = 1;i <= m;i++)
	{
		S[i] = getint(),T[i] = getint();
		dfs(S[i],0,T[i],i);
	}
	int ans = 0;
	for(int i = 1;i < n;i++) ans += std::min(cnt[i],2);
	printf("%d\n",ans);
	while(1)
	{
		int p = 0,q;
		for(int i = 1;i < n;i++) if(cnt[i] == 1 && vis[i << 1 | 1] ^ vis[i << 1]) {p = i;break;}
		if(!p) for(int i = 1;i < n;i++) if(cnt[i] >= 2) {p = i;break;}
		if(!p) break;
		for(int i = 0,sz = edge[p].size();i < sz;i++) if(!used[edge[p][i]]) {q = edge[p][i];break;}
		used[q] = 1;
		for(int i = 0,sz = path[q].size(),u = path[q][i];i < sz;u = path[q][++i])
		{
			if((u >> 1) != p) continue;
			if(vis[u])
			{
				std::swap(S[q],T[q]);
				for(int j = 0;j < sz;++j) path[q][j] ^= 1;
			}
			for(int j = 0,v = path[q][j];j < sz;v = path[q][++j]) --cnt[v >> 1],vis[v] = 1;
			break;
		}
	}
	for(int i = 1;i <= m;i++) printf("%d %d\n",S[i],T[i]);
	return 0;
}