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 std::pair;
using std::vector;
using std::string;

typedef long long ll;
typedef pair<int, int> pii;

#define fst first
#define snd second
#define pb(a) push_back(a)
#define mp(a, b) std::make_pair(a, b)
#define debug(...) fprintf(stderr, __VA_ARGS__)

template <typename T> bool chkmax(T& a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> bool chkmin(T& a, T b) { return a > b ? a = b, 1 : 0; }

const int oo = 0x3f3f3f3f;

string procStatus() {
    std::ifstream t("/proc/self/status");
    return string(std::istreambuf_iterator<char>(t), std::istreambuf_iterator<char>());
}

template <typename T> T read(T& x) {
    int f = 1; x = 0;
    char ch = getchar();
    for(;!isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
    for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return x *= f;
}

const int N = 4000;

int st[N + 5], to[N + 5], nxt[N + 5], ecnt = 1;

inline void addedge(int u, int v) {
    to[++ ecnt] = v; nxt[ecnt] = st[u]; st[u] = ecnt;
    to[++ ecnt] = u; nxt[ecnt] = st[v]; st[v] = ecnt;
}

bool get_p(int u, int fa, int t, vector<int>& p) {
    if(u == t) return true;

    for(int i = st[u]; i; i = nxt[i]) {
        int v = to[i];
        if(v == fa || !get_p(v, u, t, p)) continue;

        p.pb(i);
        return true;
    }
    return false;
}

int n, m;
int cnt[N + 5];
int x[N + 5], y[N + 5];
vector<int> p[N + 5], q[N + 5];

bool dir[N + 5], used[N + 5];

int main() {
#ifdef Wearry
    freopen("in", "r", stdin);
    freopen("out", "w", stdout);
#endif

    read(n), read(m);
    for(int i = 1; i < n; ++i) {
        static int u, v;
        read(u), read(v);
        addedge(u, v);
    }

    for(int i = 1; i <= m; ++i) {
        read(x[i]), read(y[i]);
        get_p(x[i], 0, y[i], p[i]);
        for(auto e : p[i]) ++ cnt[e >> 1], q[e >> 1].pb(i);
    }

    int ans = 0;
    for(int i = 1; i < n; ++i) ans += std::min(cnt[i], 2);

    while(true) {
        int eid = 0, pid;
        for(int i = 1; i < n; ++i) {
            if(cnt[i] == 1 && dir[i << 1] != dir[i << 1 | 1]) {
                eid = dir[i << 1] ? (i << 1 | 1) : (i << 1);
                break;
            }
        }

        if(!eid) {
            for(int i = 1; i < n; ++i) {
                if(cnt[i]) { eid = i << 1; break; }
            }
            if(!eid) break;
        }

        bool rev = 0;
        while(used[q[eid >> 1].back()]) q[eid >> 1].pop_back();

        used[pid = q[eid >> 1].back()] = true;

        for(auto e : p[pid]) if((e >> 1) == (eid >> 1)) {
            rev = e ^ eid;
        }

        if(rev) {
            std::swap(x[pid], y[pid]);
            for(auto& e : p[pid]) e ^= 1;
        }

        for(auto e : p[pid]) -- cnt[e >> 1], dir[e] = true;
    }

    printf("%d\n", ans);
    for(int i = 1; i <= m; ++i) printf("%d %d\n", x[i], y[i]);

    return 0;
}