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 = 4010;
int n, m, tot, s[maxn], t[maxn], cnt[maxn][maxn];
int ans, fa[maxn], dep[maxn], deg[maxn], dist[maxn][maxn];
bool rev[maxn], chk[maxn], chke[maxn];
vector<int> G[maxn], path[maxn];
vector<pair<int, int>> cov[maxn];

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1, u, v; i < n; i++) {
        scanf("%d %d", &u, &v), deg[u]++, deg[v]++;
        G[u].push_back(v), G[v].push_back(u);
    }
    function<void(int)> dfs = [&](int v) {
        for (int u : G[v]) if (u ^ fa[v]) {
            fa[u] = v, dep[u] = dep[v] + 1, dfs(u);
        }
    };
    dfs(1);
    for (int i = 1; i <= n; i++) {
        function<void(int, int)> dfs = [&](int v, int fa) {
            for (int u : G[v]) if (u ^ fa) {
                dist[i][u] = dist[i][v] + 1, dfs(u, v);
            }
        };
        dfs(i, 0);
    }
    auto add = [&](int id, int u, int v) {
        s[id] = u, t[id] = v;
        if (u == v) return;
        vector<int> L, R;
        while (u ^ v) {
            if (dep[u] > dep[v]) L.push_back(u), u = fa[u];
            else R.push_back(v), v = fa[v];
        }
        for (int v : L) path[id].push_back(v);
        path[id].push_back(u), reverse(R.begin(), R.end());
        for (int v : R) path[id].push_back(v);
        for (int i = 0; i < path[id].size(); i++) {
            cov[path[id][i]].push_back({id, i});
            if (i == path[id].size() - 1) continue;
            int u = path[id][i], v = path[id][i + 1];
            if (u > v) swap(u, v);
            if (cnt[u][v]++ < 2) ans++;
        }
    };
    for (int i = 1, u, v; i <= m; i++) {
        scanf("%d %d", &u, &v), add(++tot, u, v);
    }
    function<void()> solve = [&]() {
        int u = 0, v = 0;
        for (int i = 1; i <= n; i++) {
            if (!chk[i] && deg[i] == 1) { u = i; break; }
        }
        if (!u) return;
        for (int i : G[u]) {
            if (!chk[i]) { v = i; break; }
        }
        chk[u] = 1, deg[v]--;
        auto query = [&](int v, int pos) {
            return (~pos && pos < path[v].size()) ? path[v][pos] : 0;
        };
        auto get = [&](int v, int pos, int t) {
            return query(v, pos - 1) == t || query(v, pos + 1) == t;
        };
        pair<int, int> pu = {0, 0}, pv = {0, 0};
        for (auto p : cov[u]) {
            if (!chke[p.first] && get(p.first, p.second, v)) {
                if (!pu.first) pu = p;
                else if (!pv.first) pv = p;
            }
        }
        if (!pv.first) { solve(); return; }
        auto reverse = [&](int v, int x, int y) {
            if (rev[v]) swap(x, y);
            if ((dist[s[v]][x] + dist[y][t[v]] + 1) == dist[s[v]][t[v]]) rev[v] ^= 1;
        };
        reverse(pu.first, u, v), reverse(pv.first, v, u);
        chke[pu.first] = chke[pv.first] = 1;
        auto find = [&](int v, int pos, int x) {
            return query(v, pos + 1) ^ x ? path[v][0] : path[v].back();
        };
        add(++tot, find(pu.first, pu.second, v), find(pv.first, pv.second, v));
        int tmp = tot; solve();
        rev[pu.first] ^= rev[tmp], rev[pv.first] ^= rev[tmp];
    };
    printf("%d\n", ans), solve();
    for (int i = 1; i <= m; i++) {
        if (rev[i]) printf("%d %d\n", t[i], s[i]);
        else printf("%d %d\n", s[i], t[i]);
    }
    return 0;
}