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
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <cassert>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 2e3+5;
int n, m, g[N][N], d[N], a[N], b[N], u[N], v[N], f[N], tag[N], fa[N], s[N], dep[N];
void dfs(int x)
{
    for(int y = 1; y <= n; ++y) 
        if(g[x][y]&&fa[x] != y) dep[y] = dep[x]+1, fa[y] = x, dfs(y);
}
int lca(int x, int y)
{
    for( ; x != y; x = fa[x]) if(dep[x] < dep[y]) swap(x, y);
    return x;
}
int get(int x)
{
    int ret = 0;
    for(int y = 1; y <= n; ++y) 
        if(g[x][y]&&fa[x] != y) ret += get(y), s[x] += s[y];
    return ret+min(s[x], 2);
}
int main()
{
    n = read(), m = read();
    for(int i = 1; i < n; ++i)
    {
        int x = read(), y = read();
        g[x][y] = g[y][x] = 1, ++d[x], ++d[y];
    }
    dfs(1);
    for(int i = 1; i <= m; ++i) u[i] = a[i] = read(), v[i] = b[i] = read(), ++s[a[i]], ++s[b[i]], s[lca(a[i], b[i])] -= 2;
    for(int _ = 1, x, y; _ < n; ++_)
    {
        for(x = 1; x <= n&&d[x] != 1; ++x); assert(x != n+1);
        for(y = 1; y <= n&&(!d[y]||!g[x][y]); ++y); assert(y != n+1);
        int p = 0, q = 0; --d[x], --d[y];
        // printf("%d %d %d\n", _, x, y);
        for(int i = 1; i <= m; ++i)
            if((u[i] == x)^(v[i] == x)) p?q = i:p = i;
        // printf("%d %d\n", p, q);
        if(p&&q)
        {
            switch((u[p] == x)<<1|(u[q] == x))
            {
                case 3: u[p] = v[q], tag[q] = 1; break;
                case 2: u[p] = u[q]; break;
                case 1: v[p] = v[q]; break;
                case 0: v[p] = u[q], tag[q] = 1; break;
            }
            f[q] = p, u[q] = v[q];
            // printf("%d %d\n", u[p], v[p]);
        }            
        for(int i = 1; i <= m; ++i) 
            if((u[i] == x)^(v[i] == x)) u[i] == x?u[i] = y:v[i] = y;
    }
    printf("%d\n", get(1));
    for(int i = 1; i <= m; ++i) 
    {
        if(f[i]) tag[i] ^= tag[f[i]]; if(tag[i]) swap(a[i], b[i]);
        printf("%d %d\n", a[i], b[i]);
    }
    return 0;
}
/*
2 3
3 4
4 2
*/