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:
// Why I am so dumb? :c
#include <bits/stdc++.h>

#define pb push_back
#define mp make_pair

#define all(x) (x).begin(), (x).end()

#define fi first
#define se second

#define pii pair<int, int>

using namespace std;

typedef long long ll;

const int MAXN = (int)2e3 + 5;

vector<int> adj[MAXN];

int tmp[MAXN];

int n, m;

void build(int v, int pr, vector<pii> &e) {
    for (int id : adj[v]) {
        if (id == pr) {
            continue;
        }

        int to = e[id].fi + e[id].se - v;
        tmp[to] = id;
        build(to, id, e);
    }
}

void addPath(int a, int b, vector<pii> &e, vector<int> &w) {
    tmp[a] = -1;
    build(a, -1, e);

    for (int cur = b, p; tmp[cur] != -1; cur = e[p].fi + e[p].se - cur) {
        p = tmp[cur];
        ++w[p];        
    }
}

vector<pii> calc(vector<pii> e, vector<pii> arr) {
    int v = -1, par = -1;
    map<int, int> deg;

    for (pii it : e) {
        ++deg[it.fi];
        ++deg[it.se];
    }

    for (pii it : deg) {
        if (it.se == 1) {
            v = it.fi;
        }
    }

    if (v == -1) {
        return arr;
    }

    for (pii it : e) {
        if (it.fi == v || it.se == v) {
            par = it.fi + it.se - v;
        }
    }

    vector<pii> newArr, newE;
    vector<int> leaf;
    map<int, int> p;

    for (int i = 0; i < arr.size(); ++i) {
        if (arr[i].fi == arr[i].se) {
            continue;
        }

        if (arr[i].fi == v) {
            swap(arr[i].fi, arr[i].se);
        }

        if (arr[i].se == v) {
            leaf.pb(i);            
        }
        else {
            p[newArr.size()] = i;
            newArr.pb(arr[i]);
        }
    }

    vector<pair<pii, int>> dep;

    while (leaf.size() >= 2) {
        int a = leaf.back();
        leaf.pop_back();
        int b = leaf.back();
        leaf.pop_back();
        
        dep.pb(mp(mp(a, b), newArr.size()));
        newArr.pb(mp(arr[a].fi, arr[b].fi));
    }

    if (leaf.size() == 1) {
        int a = leaf.back();
        leaf.pop_back();

        dep.pb(mp(mp(a, a), newArr.size()));
        newArr.pb(mp(arr[a].fi, par));
    }

    for (pii it : e) {
        if (it.fi != v && it.se != v) {
            newE.pb(it);
        }
    }

    vector<pii> sub = calc(newE, newArr);
    vector<pii> cur = arr;

    for (pii it : p) {
        cur[it.se] = sub[it.fi];
    }

    for (auto it : dep) {
        int a = it.fi.fi, b = it.fi.se;

        if (a == b) {
            if (sub[it.se] != newArr[it.se]) {
                swap(cur[a].fi, cur[a].se);   
            }

            continue;
        }

        if (sub[it.se] == newArr[it.se]) {
            swap(cur[b].fi, cur[b].se);
        }
        else {
            swap(cur[a].fi, cur[a].se);
        }
    }

    return cur;
}

void solve() {                   
    scanf("%d %d", &n, &m);
    vector<pii> e(n - 1), arr(m);
    vector<int> w(n - 1, 0);

    for (int i = 0; i < n - 1; ++i) {
        scanf("%d %d", &e[i].fi, &e[i].se);
        adj[e[i].fi].pb(i);
        adj[e[i].se].pb(i);
    }

    for (int i = 0; i < m; ++i) {
        scanf("%d %d", &arr[i].fi, &arr[i].se);
        addPath(arr[i].fi, arr[i].se, e, w);        
    }

    int ans = 0;
                
    for (int x : w) {
        ans += min(x, 2);
    }

    printf("%d\n", ans);
    vector<pii> vv = calc(e, arr);

    for (pii it : vv) {
        printf("%d %d\n", it.fi, it.se);
    }
}

int main() {    
    int tt = 1;

    while (tt--) {
        solve();
    }

    return 0;
}