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 <cstdio>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>

using namespace std;

int f[4000000];
set <int> s[2000];
vector <pair<int, int>> v[2000];
map <pair<int, int>, int> mp;
vector <int> p[4000000];
pair <int, int> ans[4000000];

int dfs(int x, int y, int z) {
    set <int>::iterator it;
    
    if (x == y) return 1;
    
    for (it = s[x].begin(); it != s[x].end(); it++) {
        int nx = *it;
        
        if (nx != z && dfs(nx, y, x) == 1) {
            mp[make_pair(min(x, nx), max(x, nx))]++;
            return 1;
        }
    }
    
    return 0;
}

int main() {
    int n, m, c, sum = 0, i, j;
    queue <int> q;
    map <pair<int, int>, int>::iterator it;
    
    scanf("%d %d", &n, &m);
    
    for (i = 0; i < n - 1; i++) {
        int x, y;
        
        scanf("%d %d", &x, &y);
        
        x--;
        y--;
        
        s[x].insert(y);
        s[y].insert(x);
    }
    
    for (i = 0; i < m; i++) {
        int x, y;
        
        scanf("%d %d", &x, &y);
        
        x--;
        y--;
        
        v[x].push_back(make_pair(y, i));
        v[y].push_back(make_pair(x, i));
        ans[i] = make_pair(x, y);
        dfs(x, y, -1);
    }
    
    for (it = mp.begin(); it != mp.end(); it++) sum += min(it->second, 2);
    
    c = m;
    
    for (i = 0; i < n; i++) {
        if (s[i].size() == 1) q.push(i);
    }
    
    while (!q.empty()) {
        int x = q.front(), y;
        vector <pair<int, int>> w;
        
        q.pop();
        
        if (s[x].size() == 0) continue;
        y = *s[x].begin();
        s[y].erase(x);
        if (s[y].size() == 1) q.push(y);
        
        for (i = 0; i < v[x].size(); i++) {
            if (f[v[x][i].second] == 0) {
                f[v[x][i].second] = 1;
                w.push_back(v[x][i]);
            }
        }
        
        if (w.size() >= 2) {
            int nx, ny, id1, id2;
            
            nx = w.back().first;
            id1 = w.back().second;
            w.pop_back();
            
            ny = w.back().first;
            id2 = w.back().second;
            w.pop_back();
            
            if (nx == ny) {
                ans[id1] = make_pair(x, nx);
                ans[id2] = make_pair(nx, x);
            } else {
                v[nx].push_back(make_pair(ny, c));
                v[ny].push_back(make_pair(nx, c));
                p[c].push_back(id1);
                p[c].push_back(id2);
                ans[c] = make_pair(nx, ny);
                c++;
            }
        }
        
        for (i = 0; i < w.size(); i++) {
            int nx = w[i].first, ny = y, id = w[i].second;
            
            if (nx != ny) {
                v[nx].push_back(make_pair(ny, c));
                v[ny].push_back(make_pair(nx, c));
                p[c].push_back(id);
                ans[c] = make_pair(nx, ny);
                c++;
            }
        }
    }

    for (i = c - 1; i >= m; i--) {
        int x = ans[i].first, y = ans[i].second;
        
        for (j = 0; j < p[i].size(); j++) {
            int nx = ans[p[i][j]].first, ny = ans[p[i][j]].second;
            
            if (x == ny || y == nx) ans[p[i][j]] = make_pair(ny, nx);
        }
    }
    
    printf("%d\n", sum);
    
    for (i = 0; i < m; i++) printf("%d %d\n", ans[i].first + 1, ans[i].second + 1);
    
    return 0;
}