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 N = 2022;
int n, m, u[N], v[N], col[N], ans = 0;
vector<int> G[N];
vector<pair<int, int>> E[N];

int dep[N], fa[N], has[N][N], ig[N];
vector<pair<int, int>> cov[N];
inline void dfs1(int x, int pre) {
  dep[x] = dep[pre] + 1;
  fa[x] = pre;
  for (auto y : G[x]) if (y != pre) {
    dfs1(y, x);
  }
}
inline void label(int x, int y, int id) {
  while (x != y) {
    if (dep[x] >= dep[y]) {
      has[x][id] = 1;
      cov[x].emplace_back(id, 1);
      x = fa[x];
    } else {
      has[y][id] = 1;
      cov[y].emplace_back(id, 0);
      y = fa[y];
    }
  }
}
inline void dfs2(int x) {
  for (auto y : G[x]) if (y != fa[x]) {
    dfs2(y);
  }
  if (x != 1) {
    ans += (cov[x].size() >= 1) + (cov[x].size() >= 2);
    if (ig[x] || cov[x].size() < 2) return;
    int p = cov[x][0].first, q = cov[x][1].first;
    E[p].emplace_back(q, cov[x][0].second == cov[x][1].second);
    E[q].emplace_back(p, cov[x][0].second == cov[x][1].second);
    for (int y = 2; y <= n; y++) {
      if (has[y][p] && has[y][q]) {
        ig[y] = 1;
      }
    }
  }
}
inline void dfs3(int x) {
  for (auto y : E[x]) if (!~col[y.first]) {
    col[y.first] = col[x] ^ y.second;
    dfs3(y.first);
  }
}

int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  dfs1(1, 0);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u[i], &v[i]);
    label(u[i], v[i], i);
  }
  dfs2(1);
  printf("%d\n", ans);
  memset(col, -1, sizeof col);
  for (int i = 1; i <= m; i++) {
    if (!~col[i]) {
      col[i] = 0;
      dfs3(i);
    }
    if (col[i]) {
      printf("%d %d\n", v[i], u[i]);
    } else {
      printf("%d %d\n", u[i], v[i]);
    }
  }
}