#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 3e5 + 5, maxm = maxn << 1, inf = 0x7fffffff;
int x[maxn], y[maxn], z[maxn], p[maxn];//x,y,z为每条边的数据，p[x]为x代表边的权值
int head[maxm], nxt[maxm], v[maxm], cnt;//前向星
int son[maxn], dad[maxn], sz[maxn], depth[maxn], root;//树剖dfs1
int id[maxn], top[maxn], rak[maxn], num;//树剖dfs2
int c[maxn], d[maxn], srt[maxn];//记录区间并排序的数组
int ma, mb, mc;//最大路径的记录
int n, m;
struct Binary_Indexed_Tree//求和树状数组
{
int a[maxn];
}BIT;
inline int max(int x, int y) { return x > y ? x : y; }
inline int min(int x, int y) { return x < y ? x : y; }
struct Segment_Tree//最大值线段树
{
int mx[maxn << 2], tag[maxn << 2];
}ST;
inline void addline(int x, int y) { v[cnt] = y, nxt[cnt] = head[x], head[x] = cnt++; }
inline int read()
{
RG char c = getchar(); RG int x = 0;
while (c<'0' || c>'9') c = getchar();
while (c >= '0'&&c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
return x;
}
inline void dfs1(int x, int f, int d)//树剖
{
dad[x] = f, depth[x] = d, sz[x] = 1;
for (RG int i = head[x]; ~i; i = nxt[i])
{
if (v[i] == f) continue;
dfs1(v[i], x, d + 1);
sz[x] += sz[v[i]];
if (sz[v[i]] > sz[son[x]]) son[x] = v[i];
}
return;
}
inline void dfs2(int x, int t)//树剖
{
top[x] = t, id[x] = ++num, rak[id[x]] = x;
if (!son[x]) return;
dfs2(son[x], t);
for (RG int i = head[x]; ~i; i = nxt[i])
if (v[i] != dad[x] && v[i] != son[x]) dfs2(v[i], v[i]);
return;
}
inline int sum(int x, int y)//求某条路径的权值
{
RG int tx = top[x], ty = top[y], ans = 0;
while (tx != ty)
{
if (depth[tx] >= depth[ty]) ans += BIT.sum(id[tx], id[x]), x = dad[tx], tx = top[x];
else ans += BIT.sum(id[ty], id[y]), y = dad[ty], ty = top[y];
}
if (id[x] <= id[y]) ans += BIT.sum(id[x] + 1, id[y]);
else ans += BIT.sum(id[y] + 1, id[x]);
return ans;
}
inline bool cmp(int x, int y) { return c[x] < c[y]; }
inline void update(int x, int y, int z)//更新mx数组（其实是更新最大值线段树）
{
RG int tx = top[x], ty = top[y], t = 0;
while (tx != ty)
{
if (depth[tx] >= depth[ty]) c[++t] = id[tx], d[t] = id[x], x = dad[tx], tx = top[x];
else c[++t] = id[ty], d[t] = id[y], y = dad[ty], ty = top[y];
}
if (id[x] <= id[y]) c[t] = id[x] + 1, d[t] = id[y];
else c[t] = id[y] + 1, d[t] = id[x];
for (int i = 1; i <= t; i) srt[i] = i;
sort(srt + 1, srt + t + 1, cmp);
if (c[srt[1]] > 1) ST.update(1, n, 1, 1, c[srt[1]] - 1, z);
if (d[srt[t]] < n) ST.update(1, n, 1, d[srt[t]] + 1, n, z);
for (int i = 1; i < t; i) ST.update(1, n, 1, d[srt[i]] + 1, c[srt[i + 1]] - 1, z);
return;
}
inline int find_ans(int x, int y)//在最大路径上遍历并清零边求答案
{
RG int ans = inf;
if (x == y) return 0;
if (depth[x] < depth[y]) swap(x, y);
while (depth[x] != depth[y]) ans = min(ans, max(mc - p[x], ST.query(1, n, 1, id[x]))), x = dad[x];
while (x != y)
{
if (depth[x] > depth[y]) ans = min(ans, max(mc - p[x], ST.query(1, n, 1, id[x]))), x = dad[x];
else ans = min(ans, max(mc - p[y], ST.query(1, n, 1, id[y]))), y = dad[y];
}
return ans;
}
int main(void)
{
memset(head, -1, sizeof(head));
n = read(), m = read();
for (int i = 1; i < n; i++) x[i] = read(), y[i] = read(), z[i] = read();
for (int i = 1; i < n; i++) addline(x[i], y[i]), addline(y[i], x[i]);
root = rand() % n + 1, dfs1(root, 0, 1), dfs2(root, root);//树剖
for (int i = 1; i < n; i++)
{
if (depth[x[i]] > depth[y[i]]) p[x[i]] = z[i];//把深度大的点作为一条边的代表
else p[y[i]] = z[i];
}
BIT.build(n);
for (int i = 1; i <= m; i++)
{
RG int a = read(), b = read(), temp;
temp = sum(a, b), update(a, b, temp);
if (temp >= mc) ma = a, mb = b, mc = temp;//求最大路径
}
printf(*%d\n*, find_ans(ma, mb));
return 0;
}
