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 <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;
const int Maxn = 2010;
struct node {
	int y, next;
}a[Maxn<<1]; int first[Maxn], len;
void ins(int x, int y) {
	a[len].y = y;
	a[len].next = first[x]; first[x] = len;
	len++;
}
int _min(int x, int y) { return x < y ? x : y; }
int n, m;
int u[Maxn], v[Maxn];
int sum[Maxn];
bool used[Maxn], vis[Maxn<<1];
vector <int> pat[Maxn], vec[Maxn];
bool dfs(int x, int fa, int t, int p) {
	if(x == t) return true;
	for(int k = first[x]; k != -1; k = a[k].next){
		int y = a[k].y;
		if(y == fa) continue;
		if(dfs(y, x, t, p) == true){
			pat[p].push_back(k^1);
			sum[k>>1]++;
			vec[k>>1].push_back(p);
			return true;
		}
	}
	return false;
}
int main() {
	int i, j, k;
	scanf("%d%d", &n, &m);
	for(i = 1; i <= n; i++) first[i] = -1;
	for(i = 1; i < n; i++){
		int x, y;
		scanf("%d%d", &x, &y);
		ins(x, y); ins(y, x);
	}
	for(i = 1; i <= m; i++){
		scanf("%d%d", &u[i], &v[i]);
		dfs(v[i], 0, u[i], i);
	}
	int ans = 0;
	for(i = 0; i < n-1; i++) ans += _min(sum[i], 2);
	while(1){
		int id = -1;
		for(i = 0; i < n-1; i++){
			if(sum[i] == 1 && (vis[i<<1]^vis[i<<1|1]) == 1){ id = vis[i<<1] == 1 ? i<<1|1 : i<<1; break; }
		}
		if(id == -1){
			for(i = 0; i < n-1; i++) if(sum[i] > 0){ id = i<<1; break; }
			if(id == -1) break;
		}
		int p = id>>1;
		while(used[vec[p].back()]) vec[p].pop_back();
		int t = vec[p].back();
		used[vec[p].back()] = true;
		vec[p].pop_back();
		bool rev = false;
		for(i = 0; i < pat[t].size(); i++){
			if((pat[t][i]^id) == 1){ rev = true; break; }
		}
		if(rev == true){
			for(i = 0; i < pat[t].size(); i++) pat[t][i] ^= 1;
			swap(u[t], v[t]);
		}
		for(i = 0; i < pat[t].size(); i++) sum[pat[t][i]>>1]--, vis[pat[t][i]] = true;
	}
	printf("%d\n", ans);
	for(i = 1; i <= m; i++) printf("%d %d\n", u[i], v[i]);
	return 0;
}