最小化最长的路径，考虑二分答案。
问题转化成检验删去一条边的边权后，最长路径权值能否不超过 \(x\)。
考虑没删边权时，原先那些不超过 \(x\) 的路径，删去边权后肯定不会影响，直接忽略。
考虑原先比 \(x\) 长的那些路径。我们期望删边权后这些路径全部变短到 \(x\) 以内，因此称之为关键路径。
很明显，我们必须找一条被所有关键路径经过的边。否则，至少有一条关键路径的权值不会缩减，修改是无效的。
然后又很明显，我们直接贪心地找被所有关键路径经过的，权值最大的边，删除权值是最优的。
树上差分做到 \(\Theta((n + m)\log \sum t)\)。
/*
* @Author: crab-in-the-northeast
* @Date: 2023-04-19 12:55:53
* @Last Modified by: crab-in-the-northeast
* @Last Modified time: 2023-04-19 13:33:47
*/
#include <bits/stdc++.h>
inline int read() {
int x = 0;
bool f = true;
char ch = getchar();
for (; !isdigit(ch); ch = getchar())
if (ch == '-')
f = false;
for (; isdigit(ch); ch = getchar())
x = (x << 1) + (x << 3) + ch - '0';
return f ? x : (~(x - 1));
}
inline bool gmx(int &a, int b) {
return b > a ? a = b, true : false;
}
const int maxn = 300005;
const int maxm = 300005;
const int mlgn = 20;
int n, m;
typedef std :: pair <int, int> pii;
std :: vector <pii> G[maxn];
int lg[maxn], fa[maxn], prew[maxn], dep[maxn], dfn[maxn], dis[maxn], mi[mlgn][maxn], times = 0;
inline int depmin(int u, int v) {
return dep[u] < dep[v] ? u : v;
}
void dfs(int u, int ls) {
dfn[u] = ++times;
dep[u] = dep[ls] + 1;
fa[u] = mi[0][dfn[u]] = ls;
for (pii e : G[u]) {
int v = e.first, w = e.second;
if (v == ls)
continue;
dis[v] = dis[u] + w;
prew[v] = w;
dfs(v, u);
}
}
inline void getmi() {
for (int j = 1; j <= lg[n]; ++j)
for (int i = 1; i + (1 << j) - 1 <= n; ++i)
mi[j][i] = depmin(mi[j - 1][i], mi[j - 1][i + (1 << (j - 1))]);
}
inline int lca(int u, int v) {
if (u == v)
return u;
u = dfn[u];
v = dfn[v];
if (u > v)
std :: swap(u, v);
int s = lg[v - u++];
return depmin(mi[s][u], mi[s][v - (1 << s) + 1]);
}
struct query {
int s, t, L;
} q[maxn];
int cnt[maxn];
void dfs2(int u) {
for (pii e : G[u]) {
int v = e.first;
if (v == fa[u])
continue;
dfs2(v);
cnt[u] += cnt[v];
}
}
inline bool check(int x) {
int tot = 0;
std :: memset(cnt, 0, sizeof(cnt));
for (int i = 1; i <= m; ++i) {
int s = q[i].s, t = q[i].t, L = q[i].L, p = lca(s, t);
if (L <= x) {
tot = i - 1;
break;
}
++cnt[s];
++cnt[t];
cnt[p] -= 2;
}
dfs2(1);
int maxw = 0;
for (int u = 1; u <= n; ++u) {
if (cnt[u] == tot)
gmx(maxw, prew[u]);
}
return q[1].L - maxw <= x;
}
int main() {
n = read(); m = read();
for (int i = 2; i <= n; ++i)
lg[i] = lg[i >> 1] + 1;
for (int i = 1; i < n; ++i) {
int u = read(), v = read(), w = read();
G[u].push_back({v, w});
G[v].push_back({u, w});
}
dfs(1, 0);
getmi();
for (int i = 1; i <= m; ++i) {
int s = q[i].s = read(), t = q[i].t = read(), p = lca(s, t);
q[i].L = dis[s] + dis[t] - dis[p] * 2;
}
std :: sort(q + 1, q + 1 + m, [](query a, query b) {
return a.L > b.L;
});
int ans = 0;
if (check(0)) {
puts("0");
return 0;
}
for (int i = (1 << 30); i; i >>= 1)
if (!check(ans + i))
ans += i;
printf("%d\n", ans + 1);
return 0;
}