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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#define MAXN 2010
int n, m, head[MAXN], to[MAXN << 1], next[MAXN << 1], tot = 0;
inline void $(int u, int v) {
    next[tot] = head[u], to[tot] = v, head[u] = tot++;
    next[tot] = head[v], to[tot] = u, head[v] = tot++;
}
std::pair<int, int> path[MAXN];
std::vector<int> paths[MAXN], edges[MAXN];
bool dfs(int x, int y, int fa, int id) {
    if (x == y) return 1;
    for (int i = head[x]; ~i; i = next[i]) {
        if (to[i] == fa) continue;
        if (dfs(to[i], y, x, id)) {
            paths[i >> 1].push_back(id);
            edges[id].push_back(i);
            return 1;
        }
    }
    return 0;
}
int cnt[MAXN], vis[MAXN], have[MAXN << 1];
int main(int argc, char const *argv[]) {
    scanf("%d%d", &n, &m);
    memset(head, -1, sizeof(head));
    for (int i = 1, u, v; i < n; i++) scanf("%d%d", &u, &v), $(u, v);
    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &path[i].first, &path[i].second);
        dfs(path[i].first, path[i].second, 0, i);
    }

    int ans = 0;
    for (int i = 0; i < n - 1; i++) {
        cnt[i] = paths[i].size();
        ans += std::min(2, cnt[i]);
    }
    for (int p, rev;;) {
        for (p = 0; p < n - 1; p++)
            if (cnt[p] == 1 && (have[p << 1] ^ have[p << 1 | 1])) break;
        if (p == n - 1) {
            for (p = 0; p < n - 1; p++)
                if (cnt[p]) break;
            if (p == n - 1) break;
        }
        while (vis[paths[p].back()]) paths[p].pop_back();
        int x = paths[p].back();
        rev = 0;
        for (auto y : edges[x]) {
            if ((y >> 1) == p) {
                if (have[y]) {
                    rev = 1;
                    std::swap(path[x].first, path[x].second);
                }
            }
        }
        for (auto y : edges[x]) {
            have[y ^ rev] = 1;
            cnt[y >> 1]--;
        }
        vis[x] = 1;
    }
    printf("%d\n", ans);
    for (int i = 1; i <= m; i++) printf("%d %d\n", path[i].first, path[i].second);
    return 0;
}