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 X first
#define Y second
using namespace std;
int n, m, x, y, z;
vector<pair<int, int> > a[2020];
vector<int> b[2020];
vector<int> c[2020];
int d[2020], w[4020];
int st[2020], ed[2020];
bool v[2020];
bool dfs(int x, int y, int t, vector<int> &b) {
	if (x == t) {
		return true;
	}
	for (auto i: a[x]) {
		if (i.X != y) {
			b.push_back(i.Y);
			if (dfs(i.X, x, t, b)) {
				return true;
			}
			b.pop_back();
		}
	}
	return false;
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i < n; i++) {
		scanf("%d%d", &x, &y);
		a[x].push_back({y, 2 * i});
		a[y].push_back({x, 2 * i + 1});
	}
	for (int i = 0; i < m; i++) {
		scanf("%d%d", &st[i], &ed[i]);
		dfs(st[i], 0, ed[i], b[i]);
		for (int j: b[i]) {
			c[j / 2].push_back(i);
			d[j / 2]++;
		}
	}
	for (int i = 1; i < n; i++) {
		z += min(d[i], 2);
	}
	while (true) {
		int p = 0, q = -1; // p = choose edge, q = choose path
		for (int i = 1; i < n; i++) {
			if (d[i] == 1 && w[i * 2] != w[i * 2 + 1]) {
				p = i;
				break;
			}
		}
		if (p == 0) {
			for (int i = 1; i < n; i++) {
				if (d[i] >= 2) {
					p = i;
					break;
				}
			}
		}
		if (p == 0) {
			break;
		}
		for (int i: c[p]) {
			if (!v[i]) {
				q = i;
				break;
			}
		}
		v[q] = true;
		for (int i: b[q]) {
			if (i / 2 == p) {
				if (w[i]) {
					swap(st[q], ed[q]);
					for (int &j: b[q]) {
						j ^= 1;
					}
				}
				for (int &j: b[q]) {
					d[j / 2]--;
					w[j] = 1;
				}
				break;
			}
		}
	}
	printf("%d\n", z);
	for (int i = 0; i < m; i++) {
		printf("%d %d\n", st[i], ed[i]);
	}
	return 0;
}