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 N = 1e6 + 9;
  int n, m, seg[N << 2], lazy[N << 2];
  ;
  vector<pair<int, pair<int, int>>> v;
  void push(int i, int l, int r) {
    if (!lazy[i]) return;
    seg[i] += lazy[i];
    if (l != r) lazy[i << 1] += lazy[i], lazy[i << 1 | 1] += lazy[i];
    lazy[i] = 0;
  }
  void update(int i, int l, int r, int a, int b, long long int v) {
    push(i, l, r);
    if (l > b || r < a) return;
    if (l >= a && r <= b) {
      lazy[i] += v;
      return push(i, l, r);
    }
    int mid = (l + r) >> 1;
    update(i << 1, l, mid, a, b, v);
    update(i << 1 | 1, mid + 1, r, a, b, v);
    seg[i] = min(seg[i << 1], seg[i << 1 | 1]);
  }
  int get(int i, int l, int r, int a, int b) {
    push(i, l, r);
    if (l > b || r < a) return 1e9;
    if (l >= a && r <= b) return seg[i];
    int mid = (l + r) >> 1;
    int ret = get(i << 1, l, mid, a, b);
    ret = min(ret, get(i << 1 | 1, mid + 1, r, a, b));
    seg[i] = min(seg[i << 1], seg[i << 1 | 1]);
    return ret;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
      int x, y, w;
      cin >> x >> y >> w;
      v.push_back({w, {x, y}});
    }
    sort(v.begin(), v.end());
    int j = 0;
    int ans = 1e9;
    for (int i = 0; i < n; i++) {
      while (j < n && get(1, 1, m - 1, 1, m - 1) == 0) {
        update(1, 1, m - 1, v[j].second.first, v[j].second.second - 1, 1);
        j++;
      }
      if (get(1, 1, m - 1, 1, m - 1) != 0) {
        ans = min(ans, v[j - 1].first - v[i].first);
      }
      update(1, 1, m - 1, v[i].second.first, v[i].second.second - 1, -1);
    }
    cout << ans << '\n';
  }
  