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("Ofast")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  long long t[10000000], tmin[10000000];
  void add(int l, int r, int lx, int rx, int zn, int v) {
    if (l > rx || r < lx) return;
    if (l >= lx && r <= rx) {
      t[v] += zn;
      tmin[v] = min(tmin[v * 2 + 1], tmin[v * 2 + 2]) + t[v];
      return;
    }
    add(l, (l + r) / 2, lx, rx, zn, v * 2 + 1);
    add((l + r) / 2 + 1, r, lx, rx, zn, v * 2 + 2);
    tmin[v] = min(tmin[v * 2 + 1], tmin[v * 2 + 2]) + t[v];
  }
  int main() {
    ios::sync_with_stdio(0);
    int n, m;
    cin >> n >> m;
    vector<pair<long long, pair<long long, long long> > > a(n);
    for (int i = 0; i < n; i++)
      cin >> a[i].second.first >> a[i].second.second >> a[i].first;
    sort(a.begin(), a.end());
    for (int i = 0; i < 10000000; i++) {
      t[i] = 0;
      tmin[i] = 0;
    }
    for (int i = 0; i < n; i++) {
      if (a[i].second.first == 1) a[i].second.first--;
      a[i].second.first++;
    }
    int l = -1;
    long long res = 1e18;
    for (int i = 0; i < n; i++) {
      while (tmin[0] == 0) {
        if (l == n) break;
        l++;
        if (l < n)
          add(1, m, a[l].second.first, a[l].second.second, 1, 0);
        else
          break;
      }
      if (tmin[0] > 0) res = min(res, a[l].first - a[i].first);
      add(1, m, a[i].second.first, a[i].second.second, -1, 0);
    }
    cout << res << "\n";
  }
  