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>
  using namespace std;
  const int maxn = 1e6 + 10;
  int seg[maxn << 2], laz[maxn << 2], ord[maxn], le[maxn], ri[maxn], we[maxn];
  int n, m;
  void add(int val, int st, int en, int l = 0, int r = m, int id = 1) {
    if (st >= r or en <= l) return;
    if (st <= l and r <= en) {
      seg[id] += val;
      laz[id] += val;
      return;
    }
    int mid = (l + r) >> 1;
    add(val, st, en, l, mid, id << 1 | 0);
    add(val, st, en, mid, r, id << 1 | 1);
    seg[id] = min(seg[id << 1 | 0], seg[id << 1 | 1]) + laz[id];
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    m--;
    for (int i = 0; i < n; i++) {
      cin >> le[i] >> ri[i] >> we[i];
      le[i]--, ri[i]--;
    }
    iota(ord, ord + n, 0);
    sort(ord, ord + n, [](int i, int j) -> bool { return we[i] < we[j]; });
    int ans = 1e6;
    int ptr = 0;
    for (int i = 0; i < n; i++) {
      add(+1, le[ord[i]], ri[ord[i]]);
      while (seg[1] > 0 and ptr <= i) {
        add(-1, le[ord[ptr]], ri[ord[ptr]]);
        ptr++;
      }
      if (ptr - 1 >= 0) (ans) = min((ans), (we[ord[i]] - we[ord[ptr - 1]]));
    }
    cout << ans << '\n';
    return 0;
  }
  