---

首先有朴素的\(O(n^2)\)想法

首先枚举断边，之后对于断边之后的两棵子树求出直径

考虑优化这个朴素的想法

考虑换根\(dp\)

具体而言，首先求出\(f[i], fs[i]\)表示\(i\)号点向下的最长链以及\(i\)号子树内部最长的直径

并且在求出\(g[i]\)表示\(fa[i]\)在\(i\)号节点子树外的最长链

\(gs[i]\)表示\(i\)号节点子树外的直径

对于所有的\(fs[i] * gs[i]\)取\(max\)即为答案

---

首先一遍\(dfs\)把\(f, fs\)求出来

考虑怎么求\(gs[o]\)，有以下几种可能

• 一条\(fa[o]\)引申出去的链 + \(fa[o]\) 除了\(o\)子树以外的最长链

• \(gs[fa]\)

• 两条\(fa[o]\)除了\(o\)子树以外的链的和

用前缀和后缀来做到删除

---

#include <cstdio>

#include <cstring>

#include <iostream>

using namespace std;

#define ll long long

#define ri register int

#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)

#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

#define tpr template <typename ra>

tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }

tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; }

#define gc getchar

inline int read() {

    int p = 0, w = 1; char c = gc();

    while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }

    while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();

    return p * w;

}

const int sid = 200050;

ll ans;

int n, cnp;

int cap[sid], nxt[sid], node[sid];

inline void addedge(int u, int v) {

    nxt[++ cnp] = cap[u]; cap[u] = cnp; node[cnp] = v;

}

int f[sid], fs[sid];

int g[sid], gs[sid], q[sid], ps[sid], ss[sid];

#define cur node[i]

void dfs1(int o, int fa) {

    for(int i = cap[o]; i; i = nxt[i])

    if(cur != fa) {

        dfs1(cur, o);

        cmax(fs[o], fs[cur]);

        cmax(fs[o], f[cur] + f[o] + 1);

        cmax(f[o], f[cur] + 1);

    }

}

void dfs2(int o, int fa) {

    if(!fa) g[o] = -1;

    int tot = 0;

    int sx = 0, mx = 0, px = 0;

    for(int i = cap[o]; i; i = nxt[i])

        if(cur != fa) q[++ tot] = cur;

    rep(i, 1, tot + 1) ps[i] = ss[i] = 0;

    rep(i, 1, tot) {

        int p = q[i];

        cmax(g[p], g[o] + 1);

        cmax(g[p], ps[i - 1]);

        ps[i] = max(ps[i - 1], f[p] + 1);

    }

    drep(i, tot, 1) {

        int p = q[i];

        cmax(g[p], ss[i +  1]);

        ss[i] = max(ss[i + 1], f[p] + 1);

    }

    rep(i, 1, tot) {

        int p = q[i];

        cmax(gs[p], ps[i - 1] + ss[i + 1]);

        cmax(gs[p], gs[o]);

        cmax(gs[p], g[o] + 1 + ps[i - 1]);

        cmax(gs[p], g[o] + 1 + ss[i + 1]);

    }

    rep(i, 1, tot) {

        int p = q[i];

        cmax(gs[p], mx + sx);

        cmax(gs[p], px); cmax(px, fs[p]);

        if(f[p] + 1 >= mx) sx = mx, mx = f[p] + 1;

        else if(f[p] + 1 >= sx) sx = f[p] + 1;

    }

    mx = sx = px = 0;

    drep(i, tot, 1) {

        int p = q[i];

        cmax(gs[p], mx + sx);

        cmax(gs[p], px); cmax(px, fs[p]);

        if(f[p] + 1 >= mx) sx = mx, mx = f[p] + 1;

        else if(f[p] + 1 >= sx) sx = f[p] + 1;

    }

    for(int i = cap[o]; i; i = nxt[i])

        if(cur != fa) {

            cmax(ans, 1ll * fs[cur] * gs[cur]);

            dfs2(cur, o);

        }

}

int main() {

    n = read();

    rep(i, 2, n) {

        int u = read(), v =read();

        addedge(u, v); addedge(v, u);

    }

    dfs1(1, 0); dfs2(1, 0);

    printf("%lld\n", ans);

    return 0;

}