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:
//2018-6-7
//miaowey
//
#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define For(i, a, b) for(int i = (a); i <= (int)(b); ++i)

#define N (4000 + 5)

int n, m, eu[N], ev[N], cnt[N];
int en, Begin[N], Nxt[N], To[N];

bool flag[N], use[N];
vector<int> P[N], cov[N];

inline void Add(int u, int v){
	To[++en] = v; Nxt[en] = Begin[u]; Begin[u] = en;
}

bool Dfs(int u, int F, int aim, vector<int> &s){
	if(u == aim) return true;
	
	for(int i = Begin[u]; i; i = Nxt[i]){
		int v = To[i];
		if(v == F) continue;
		if(Dfs(v, u, aim, s)) {s.pb(i); return true;}
	}

	return false;
}

int main(){	
	int u, v, ans = 0;
	en = 1;

	scanf("%d%d", &n, &m);
	For(i, 1, n - 1){
		scanf("%d%d", &u, &v);
		Add(u, v), Add(v, u);
	}

	For(i, 1, m){
		scanf("%d%d", &u, &v);
		eu[i] = u, ev[i] = v;

		Dfs(u, 0, v, P[i]);
		
		for(int t : P[i]) ++cnt[t >> 1], cov[t >> 1].pb(i);
	}

	For(i, 1, n - 1) ans += min(cnt[i], 2);
	
	while(true){
	
		int id = 0;
		For(i, 1, n - 1) if(cnt[i] == 1 && (flag[i << 1] ^ flag[i << 1 | 1])){
			id = flag[i << 1]? (i << 1 | 1): (i << 1);
			break;
		}

		if(!id){
			For(i, 1, n - 1) if(cnt[i]) id = i << 1;
			if(!id) break;
		}

		int o = id >> 1;
		while(use[cov[o].back()]) cov[o].pop_back();
		
		int y = cov[o].back();
		use[y] = true;
		
		bool rev = false;
		for(int t : P[y]) if((t >> 1) == o){
			if(id ^ t) rev = true;
			break;
		}

		if(rev){
			swap(eu[y], ev[y]);
			for(int &t : P[y]) t ^= 1;
		}

		for(int t : P[y]) --cnt[t >> 1], flag[t] = true;
	}

	printf("%d\n", ans);
	For(i, 1, m) printf("%d %d\n", eu[i], ev[i]);

	return 0;
}