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 fir first
#define sec second
using namespace std;
typedef pair<int, int> pii;
const int maxn = 2005;

int n, m, deg[maxn];

int head[maxn], tot;
struct edge{
	int nxt, to;
}e[maxn<<1];

void Addedge(int x, int y)
{
	e[++tot] = (edge){head[x], y};
	head[x] = tot;
}

int f[maxn], dep[maxn], s[maxn];

void dfs1(int x)
{
	int id;
	for(int i = head[x]; i; i = e[i].nxt)
	{
		id = e[i].to;
		if(id == f[x])	continue;
		f[id] = x;
		dep[id] = dep[x] + 1;
		dfs1(id);
	}
}

int lca(int x, int y)
{
	if(dep[x] < dep[y])	swap(x, y);
	while(dep[x] > dep[y])
		x = f[x];
	while(x != y)
	{
		x = f[x];
		y = f[y];
	}
	return x;
}

void dfs2(int x)
{
	int id;
	for(int i = head[x]; i; i = e[i].nxt)
	{
		id = e[i].to;
		if(id == f[x])	continue;
		dfs2(id);
		s[x] += s[id];
	}
}

pii a[maxn], b[maxn];
int col[maxn], mat[maxn];
bool vis[maxn];

int main()
{
	scanf("%d%d", &n, &m);
	int u, v, id1, id2, ans = 0;
	for(int i = 1; i < n; ++i)
	{
		scanf("%d%d", &u, &v);
		Addedge(u, v);
		Addedge(v, u);
		++ deg[u];
		++ deg[v];
	}
	dfs1(1);
	for(int i = 1; i <= m; ++i)
	{
		scanf("%d%d", &u, &v);
		a[i] = b[i] = make_pair(u, v);
		++ s[u];
		++ s[v];
		s[lca(u, v)] -= 2;
	}
	dfs2(1);
	for(int i = 1; i <= n; ++i)
		ans += min(s[i], 2);
	printf("%d\n", ans);
	for(int i = 1; i < n; ++i)
	{
		for(int j = 1; j <= n; ++j)
			if(deg[j] == 1 && !vis[j])
			{
				u = j;
				break;
			}
		vis[u] = 1;
		for(int j = head[u]; j; j = e[j].nxt)
		{
			v = e[j].to;
			if(!vis[v])	break;
		}
		-- deg[v];
		id1 = id2 = 0;
		for(int j = 1; j <= m; ++j)
			if((b[j].fir == u) ^ (b[j].sec == u))
			{
				if(!id1)
				{
					id1 = j;
				}
				else
				{
					id2 = j;
					break;
				}
			}
		if(!id1)	continue;
		if(!id2)
		{
			if(b[id1].fir == u)
				b[id1].fir = v;
			else
				b[id1].sec = v;
			continue;
		}
		if(b[id1].fir == u)
		{
			if(b[id2].fir == u)
			{
				b[id1].fir = b[id2].sec;
				col[id2] = 1;
			}
			else
				b[id1].fir = b[id2].fir;
		}
		else
		{
			if(b[id2].sec == u)
			{
				b[id1].sec = b[id2].fir;
				col[id2] = 1;
			}
			else
				b[id1].sec = b[id2].sec;
		}
		b[id2].fir = b[id2].sec;
		mat[id2] = id1;
		for(int j = 1; j <= m; ++j)
			if((b[j].fir == u) ^ (b[j].sec == u))
			{
				if(b[j].fir == u)
					b[j].fir = v;
				else
					b[j].sec = v;
			}
	}
	for(int i = 1; i <= m; ++i)
		if(mat[i])
		{
			col[i] ^= col[mat[i]];
		}
	for(int i = 1; i <= m; ++i)
		printf("%d %d\n", col[i]? a[i].sec: a[i].fir, col[i]? a[i].fir: a[i].sec);
	return 0;
}