description: You are given n segments on a number line, numbered from 1 to n. The i-th segments covers all integer points from l_i to r_i and has a value w_i.
  
  You are asked to select a subset of these segments (possibly, all of them). Once the subset is selected, it's possible to travel between two integer points if there exists a selected segment that covers both of them. A subset is good if it's possible to reach point m starting from point 1 in arbitrary number of moves.
  
  The cost of the subset is the difference between the maximum and the minimum values of segments in it. Find the minimum cost of a good subset.
  
  In every test there exists at least one good subset.
  
  Input
  
  The first line contains two integers n and m (1 ≤ n ≤ 3 ⋅ 10^5; 2 ≤ m ≤ 10^6) — the number of segments and the number of integer points.
  
  Each of the next n lines contains three integers l_i, r_i and w_i (1 ≤ l_i < r_i ≤ m; 1 ≤ w_i ≤ 10^6) — the description of the i-th segment.
  
  In every test there exists at least one good subset.
  
  Output
  
  Print a single integer — the minimum cost of a good subset.
  
  Examples
  
  Input
  
  
  5 12
  1 5 5
  3 4 10
  4 10 6
  11 12 5
  10 12 3
  
  
  Output
  
  
  3
  
  
  Input
  
  
  1 10
  1 10 23
  
  
  Output
  
  
  0
solution: #include <bits/stdc++.h>
  #pragma GCC optimize(3)
  using namespace std;
  const int INF = 0x3f3f3f3f;
  const int N = 1e6 + 10;
  const long long mod = 1000000007;
  const double eps = 1e-9;
  const double PI = acos(-1);
  template <typename T>
  inline void read(T &a) {
    char c = getchar();
    T x = 0, f = 1;
    while (!isdigit(c)) {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (isdigit(c)) {
      x = (x << 1) + (x << 3) + c - '0';
      c = getchar();
    }
    a = f * x;
  }
  int gcd(int a, int b) { return (b > 0) ? gcd(b, a % b) : a; }
  struct P {
    int l, r, w;
  } a[N];
  struct Node {
    int l, r;
    int x, add;
  } tr[N * 4];
  int n, m;
  bool cmp(P A, P B) { return A.w < B.w; }
  void push(int u) { tr[u].x = min(tr[u << 1].x, tr[u << 1 | 1].x); }
  void down(int u) {
    if (tr[u].add != 0) {
      int &k = tr[u].add;
      tr[u << 1].x += k;
      tr[u << 1 | 1].x += k;
      tr[u << 1].add += k;
      tr[u << 1 | 1].add += k;
      k = 0;
    }
  }
  void build(int u, int l, int r) {
    tr[u] = {l, r, 0, 0};
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
  }
  void modify(int u, int l, int r, int k) {
    if (tr[u].l >= l && tr[u].r <= r) {
      tr[u].x += k, tr[u].add += k;
      return;
    }
    down(u);
    int mid = (tr[u].l + tr[u].r) >> 1;
    if (l <= mid) modify(u << 1, l, r, k);
    if (r > mid) modify(u << 1 | 1, l, r, k);
    push(u);
  }
  int main() {
    read(n), read(m);
    m--;
    build(1, 1, m);
    for (int i = 1; i <= n; i++) read(a[i].l), read(a[i].r), read(a[i].w);
    sort(a + 1, a + 1 + n, cmp);
    int l = 0, r = 0, ans = INF;
    while (r <= n) {
      while (r + 1 <= n && tr[1].x == 0) r++, modify(1, a[r].l, a[r].r - 1, 1);
      if (!tr[1].x) break;
      while (l + 1 <= r && tr[1].x > 0) l++, modify(1, a[l].l, a[l].r - 1, -1);
      ans = min(ans, a[r].w - a[l].w);
    }
    printf("%d\n", ans);
    ;
    return 0;
  }
  