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>
using namespace std;
const int MAXN = 5e3 + 5;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
vector <int> a[MAXN];
int depth[MAXN], father[MAXN];
int n, m, tot, ans, x[MAXN], y[MAXN];
bool used[MAXN], ok[MAXN], res[MAXN];
int nowx[MAXN], nowy[MAXN], f[MAXN], g[MAXN];
void dfs(int pos, int fa) {
	depth[pos] = depth[fa] + 1;
	father[pos] = fa;
	for (auto x : a[pos])
		if (x != fa) dfs(x, pos);
}
void mark(int a, int b, int c, int d) {
	static int cnt[MAXN];
	memset(cnt, 0, sizeof(cnt));
	while (a != b) {
		if (depth[a] < depth[b]) swap(a, b);
		cnt[a]++, a = father[a];
	}
	while (c != d) {
		if (depth[c] < depth[d]) swap(c, d);
		cnt[c]++, c = father[c];
	}
	for (int i = 1; i <= n; i++)
		ok[i] |= cnt[i] >= 2;
}
void work(int pos, int fa) {
	for (auto x : a[pos])
		if (x != fa) work(x, pos);
	if (pos != 1) {
		int cnt = 0;
		for (int i = 1; i <= tot; i++)
			if (!used[i]) cnt += (nowx[i] == pos) ^ (nowy[i] == pos);
		ans += min(max(cnt, ok[pos] * 2), 2);
		if (cnt <= 1 || ok[pos]) {
			for (int i = 1; i <= tot; i++)
				if (!used[i]) {
					if (nowx[i] == pos) nowx[i] = fa;
					if (nowy[i] == pos) nowy[i] = fa;
				}
			return;
		}
		vector <int> points;
		for (int i = 1; i <= tot; i++)
			if (!used[i] && ((nowx[i] == pos) ^ (nowy[i] == pos))) {
				points.push_back(i);
			}
		int a = points[0], b = points[1];
		mark(nowx[a], nowy[a], nowx[b], nowy[b]);
		used[a] = used[b] = true;
		if (nowx[a] == pos && nowx[b] == pos) {
			tot++, f[tot] = a, g[tot] = b;
			x[tot] = y[a], nowx[tot] = nowy[a];
			y[tot] = y[b], nowy[tot] = nowy[b];
		}
		if (nowy[a] == pos && nowx[b] == pos) {
			tot++, f[tot] = a, g[tot] = b;
			x[tot] = x[a], nowx[tot] = nowx[a];
			y[tot] = y[b], nowy[tot] = nowy[b];
		}
		if (nowx[a] == pos && nowy[b] == pos) {
			tot++, f[tot] = a, g[tot] = b;
			x[tot] = y[a], nowx[tot] = nowy[a];
			y[tot] = x[b], nowy[tot] = nowx[b];
		}
		if (nowy[a] == pos && nowy[b] == pos) {
			tot++, f[tot] = a, g[tot] = b;
			x[tot] = x[a], nowx[tot] = nowx[a];
			y[tot] = x[b], nowy[tot] = nowx[b];
		}
		for (int i = 1; i <= tot; i++)
			if (!used[i]) {
				if (nowx[i] == pos) nowx[i] = fa;
				if (nowy[i] == pos) nowy[i] = fa;
			}
	}
}
int main() {
	read(n), read(m), tot = m;
	for (int i = 1; i <= n - 1; i++) {
		int x, y; read(x), read(y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	for (int i = 1; i <= m; i++) {
		read(x[i]), read(y[i]);
		nowx[i] = x[i];
		nowy[i] = y[i];
	}
	dfs(1, 0);
	work(1, 0);
	writeln(ans);
	for (int i = tot; i >= 1; i--) {
		if (!used[i]) res[i] = true;
		if (i > m) {
			int a = abs(f[i]), b = abs(g[i]);
			res[a] = res[i] ^ (x[i] != x[a]);
			res[b] = res[i] ^ (y[i] != y[b]);
		}
	}
	for (int i = 1; i <= m; i++)
		if (res[i]) printf("%d %d\n", x[i], y[i]);
		else printf("%d %d\n", y[i], x[i]);
	return 0;
}