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;
constexpr int N = 2E3;
int n, m, ans;
vector<int> e[N], p[N], de, di;
vector<pair<int, int>> path;
int parent[N], dep[N], c[N], in[N];
int dfs_now;
void dfs(int u) {
    in[u] = dfs_now++;
    if (parent[u] != -1)
        e[u].erase(find(e[u].begin(), e[u].end(), parent[u]));
    for (int v : e[u]) {
        dep[v] = dep[u] + 1;
        parent[v] = u;
        dfs(v);
    }
}
int lca(int u, int v) {
    while (u != v) {
        if (dep[u] < dep[v])
            swap(u, v);
        u = parent[u];
    }
    return u;
}
void dfs1(int u) {
    reverse(e[u].begin(), e[u].end());
    for (int v : e[u]) {
        dfs1(v);
        c[u] += c[v];
    }
    ans += min(c[u], 2);
    if ((int)p[u].size() >= 2) {
        int i = p[u].back();
        p[u].pop_back();
        int j = p[u].back();
        p[u].pop_back();
        int x = path[i].first;
        int y = path[j].first;
        if (x != y) {
            if (in[x] > in[y]) {
                swap(x, y);
                swap(i, j);
            }
            de[i] = de[j] = (int)path.size();
            p[y].push_back((int)path.size());
            path.emplace_back(x, y);
            de.push_back(-1);
            di.push_back(0);
        }
        di[j] = 1;
    }
    while (!p[u].empty()) {
        int i = p[u].back();
        p[u].pop_back();
        int x = path[i].first;
        int y = parent[u];
        if (x == y)
            continue;
        if (in[x] > in[y]) {
            swap(x, y);
            di[i] = 1;
        }
        de[i] = (int)path.size();
        p[y].push_back((int)path.size());
        path.emplace_back(x, y);
        de.push_back(-1);
        di.push_back(0);
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m;
    for (int i = 0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        e[--u].push_back(--v);
        e[v].push_back(u);
    }
    parent[0] = -1;
    dfs(0);
    path.resize(m);
    de.assign(m, -1);
    di.resize(m);
    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        int l = lca(--u, --v);
        ++c[u];
        ++c[v];
        c[l] -= 2;
        if (in[u] > in[v])
            swap(u, v);
        p[v].push_back(i);
        path[i] = {u, v};
    }
    dfs1(0);
    for (int i = (int)path.size() - 1; i >= 0; --i)
        if (de[i] != -1)
            di[i] ^= di[de[i]];
    for (int i = 0; i < m; ++i)
        if (di[i] == 1)
            swap(path[i].first, path[i].second);
    cout << ans << "\n";
    for (int i = 0; i < m; ++i) {
        int u, v;
        tie(u, v) = path[i];
        cout << ++u << " " << ++v << "\n";
    }
    return 0;
}