[CF468D] Tree

Little X has a tree consisting of n nodes (they are numbered from 1 to n). Each edge of the tree has a positive length. Let's define the distance between two nodes v and u (we'll denote it d(v, u)) as the sum of the lengths of edges in the shortest path between v and u.

A permutation p is a sequence of n distinct integers p₁, p₂, ..., p_n (1 ≤ p_i ≤ n). Little X wants to find a permutation p such that sum $\text{[math]}$ is maximal possible. If there are multiple optimal permutations, he wants to find the lexicographically smallest one. Help him with the task!

Input

The first line contains an integer n (1 ≤ n ≤ 10⁵).

Each of the next n - 1 lines contains three space separated integers u_i, v_i, w_i (1 ≤ u_i, v_i ≤ n; 1 ≤ w_i ≤ 10⁵), denoting an edge between nodes u_i and v_i with length equal to w_i.

It is guaranteed that these edges form a tree.

Output

In the first line print the maximum possible value of the described sum. In the second line print n integers, representing the lexicographically smallest permutation.

Examples

Input

Copy

2
1 2 3

Output

Copy

6
2 1

Input

Copy

Output

Copy

32
2 1 4 5 3



写了一上午，我太辣鸡了。
 $\sum d i s (i, p$ 
前面一定是不变的， 后面我们可以把它降低到0， 我们选择树的重心作为根，可以把后边变成0.
第一问解决。把一个点看成两个， 一个是入点，一个是出点.
记为in[x], out[x];要满足子树x可以匹配到答案， 我们必须让 $\large in[x] + out[x] <= \sum in[i]$

$in[x]$
$ f[i, j] = $
$\LARGE f[i, j]=\frac{\sum_{to}^{ } f[nxt[nxt[i,j],j],to] + f[nxt[nxt[i,j],j],j]}{deg[j]+1} + 1$
latex炸了不写了

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <set>
using namespace std;
inline int read() {
    int res = 0;char ch=getchar();
    while(!isdigit(ch)) ch=getchar();
    while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch^48), ch=getchar();
    return res;
}
#define reg register
#define N 100005
#define ll long long

int n;
struct edge {
    int nxt, to, val;
}ed[N*2];int head[N], cnt;
inline void add(int x, int y, int z) {
    ed[++cnt] = (edge) {head[x], y, z};
    head[x] = cnt;
}
ll ans;
int siz[N], root, mrt = 1e9;

void dfs(int x, int fa) 
{
    siz[x] = 1;
    int tmp = 0;
    for (reg int i = head[x] ; i ; i = ed[i].nxt)
    {
        int to = ed[i].to;
        if (to == fa) continue;
        dfs(to, x);
        siz[x] += siz[to];
        tmp = max(tmp, siz[to]);
    }
    tmp = max(tmp, n - siz[x]);
    if (tmp < mrt) mrt = tmp, root = x;
}

void gfs(int x, int fa, ll d) {
    ans += d;
    for (reg int i = head[x] ; i ; i = ed[i].nxt) 
    {
        int to = ed[i].to;
        if (to == fa) continue;
        gfs(to, x, d + ed[i].val);
    }
}

set <int> all;
set <int> tr[N];
set < pair <int, int> > inout;
#define makp(a, b) make_pair(a, b)

int belong[N], sum[N];
void ffs(int x, int fa, int bel) {
    belong[x] = bel;
    tr[bel].insert(x);
    siz[x] = 1;
    for (reg int i = head[x] ; i ; i = ed[i].nxt) 
    {
        int to = ed[i].to;
        if (to == fa) continue;
        ffs(to, x, bel);
        siz[x] += siz[to];
    }
}

inline int Find(int x) 
{
    set < pair <int, int> > :: iterator it = inout.lower_bound(makp(n - x + 1, 0));
    bool must = 0;
    if (it != inout.end() and (*it).first == n - x + 1 and (*it).second != belong[x])  must = 1;
    if (belong[x]) {
        inout.erase(makp(sum[belong[x]], belong[x]));
        sum[belong[x]]--;
        inout.insert(makp(sum[belong[x]], belong[x]));
    }
    if (must) {
        int k = *tr[(*it).second].begin();
        all.erase(k);
        tr[(*it).second].erase(k);
        if (tr[(*it).second].size()) all.insert(*tr[(*it).second].begin());
        inout.erase(makp(sum[belong[k]], belong[k]));
        inout.insert(makp(--sum[belong[k]], belong[k]));
        return k;        
    } else {
        set <int> :: iterator itt = all.begin();
        if(belong[*itt] == belong[x] and x != root) itt++;
        all.erase(itt);
        int k = *itt;
        if (belong[k]) {
            tr[belong[k]].erase(k);
            if (tr[belong[k]].size()) all.insert(*tr[belong[k]].begin());
            inout.erase(makp(sum[belong[k]], belong[k]));
            inout.insert(makp(--sum[belong[k]], belong[k]));        
        }
        return k;
    }
    return 0;
}

int main()
{
    n = read();
    for (reg int i = 1 ; i < n ; i ++) {
        int x = read(), y = read(), z = read();
        add(x, y, z), add(y, x, z);
    }
    
    root = 1;
    dfs(1, 0);
    ans = 0;
    gfs(root, 0, 0);
    printf("%lld\n", ans * 2);
    all.insert(root);
    
    for (reg int i = head[root] ; i ; i = ed[i].nxt) 
    {
        int to = ed[i].to;
        ffs(to, root, to);
        sum[to] = siz[to] * 2;
        inout.insert(makp(sum[to], to));
        siz[root] += siz[to];
        all.insert(*tr[to].begin());
    }
    
    for (reg int i = 1 ; i <= n ; i ++)
        printf("%d ", Find(i));
    return 0;
}

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