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;

using int64 = long long;

int N, M;
vector< int > g[2000];
int in[2000];
set< pair< int, int > > q[2000];
pair< int, int > ans[4000];
int ptr;

void dfs(int rest) {
  if(rest == 1) {
    return;
  }
  int leaf = -1;
  for(int i = 0; i < N; i++) {
    if(in[i] == 1) leaf = i;
  }
  assert(leaf >= 0);
  int other = -1;
  for(auto &to : g[leaf]) {
    if(in[to] == 0) continue;
    other = to;
  }
  assert(other >= 0);
  --in[other];
  --in[leaf];
  if(q[leaf].size() == 0) {
    dfs(rest - 1);
  } else if(q[leaf].size() == 1) {
    auto s = q[leaf].begin();
    q[s->first].erase({leaf, s->second});
    if(s->first != other) {
      q[other].emplace(*s);
      q[s->first].emplace(other, s->second);
    }
    dfs(rest - 1);
    if(s->first == other) {
      ans[s->second] = {leaf, other};
    } else {
      if(ans[s->second].first == other) ans[s->second].first = leaf;
      else ans[s->second].second = leaf;
      q[s->first].erase({other, s->second});
      q[other].erase(*s);
    }
    q[s->first].emplace(leaf, s->second);
  } else {
    auto s = q[leaf].begin();
    auto t = next(s);
    q[s->first].erase({leaf, s->second});
    q[t->first].erase({leaf, t->second});

    if(s->first != other || t->first != other) {
      if(s->first != t->first) {
        q[s->first].emplace(t->first, ptr);
        q[t->first].emplace(s->first, ptr);
      } else {
        q[other].emplace(t->first, ptr);
        q[t->first].emplace(other, ptr);
      }
    }

    auto it = next(t);
    while(it != q[leaf].end()) {
      if(other != it->first) {
        q[it->first].erase({leaf, it->second});
        q[it->first].emplace(other, it->second);
        q[other].emplace(*it);
      } else {
        q[it->first].erase({leaf, it->second});
      }
      ++it;
    }
    ++ptr;
    dfs(rest - 1);

    --ptr;

    it = next(t);
    while(it != q[leaf].end()) {
      if(other != it->first) {
        if(ans[it->second].first == other) ans[it->second].first = leaf;
        else ans[it->second].second = leaf;
        q[it->first].emplace(leaf, it->second);
        q[it->first].erase({other, it->second});
        q[other].emplace(*it);
      } else {
        q[it->first].emplace(leaf, it->second);
        ans[it->second] = {leaf, other};
      }
      ++it;
    }

    if(s->first != other || t->first != other) {
      if(s->first != t->first) {
        if(ans[ptr].first != s->first) swap(s, t);
        ans[s->second] = {ans[ptr].first, leaf};
        ans[t->second] = {leaf, ans[ptr].second};
        q[s->first].erase({t->first, ptr});
        q[t->first].erase({s->first, ptr});
      } else {
        q[other].erase({t->first, ptr});
        q[t->first].erase({other, ptr});
        if(ans[ptr].first == other) ans[s->second] = {leaf, ans[ptr].second};
        else ans[s->second] = {ans[ptr].first, leaf};
        ans[t->second] = ans[s->second];
        swap(ans[t->second].first, ans[t->second].second);
      }
    } else {
      ans[s->second] = {leaf, other};
      ans[t->second] = {other, leaf};
    }
    q[s->first].emplace(leaf, s->second);
    q[t->first].emplace(leaf, t->second);
  }
  ++in[other];
  ++in[leaf];
}

int cross[2002][2002];

bool add_path(int s, int t, int p = -1) {
  if(s == t) return true;
  for(auto &to : g[s]) {
    if(to == p) continue;
    if(add_path(to, t, s)) {
      cross[min(s, to)][max(to, s)]++;
      return true;
    }
  }
  return false;
}

int main() {
  cin >> N >> M;
  for(int i = 1; i < N; i++) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    g[x].emplace_back(y);
    g[y].emplace_back(x);
    in[x]++;
    in[y]++;
  }
  for(int i = 0; i < M; i++) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    add_path(x, y);
    q[x].emplace(y, i);
    q[y].emplace(x, i);
  }
  ptr = M;
  dfs(N);
  int ret = 0;
  for(int i = 0; i < 2000; i++) {
    for(int j = 0; j < 2000; j++) {
      ret += min(2, cross[i][j]);
    }
  }
  cout << ret << endl;
  for(int i = 0; i < M; i++) {
    cout << ans[i].first + 1 << " " << ans[i].second + 1 << endl;
  }
}