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 + 5;
  struct Seg {
    int l, r, w;
    bool operator<(const Seg& t) const { return w < t.w; }
  } seg[N];
  int n, m;
  struct Node {
    int l, r;
    int x, lz;
  } tr[N << 2];
  void pushup(int u) { tr[u].x = min(tr[u << 1].x, tr[u << 1 | 1].x); }
  void pushdown(int u) {
    auto &root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
    if (root.lz) {
      left.x += root.lz;
      left.lz += root.lz;
      right.x += root.lz;
      right.lz += root.lz;
      root.lz = 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);
    pushup(u);
  }
  void add(int u, int l, int r, int x) {
    if (l <= tr[u].l && r >= tr[u].r) {
      tr[u].x += x;
      tr[u].lz += x;
      return;
    }
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid) add(u << 1, l, r, x);
    if (r > mid) add(u << 1 | 1, l, r, x);
    pushup(u);
  }
  int query(int u, int l, int r) {
    if (l <= tr[u].l && r >= tr[u].r) return tr[u].x;
    pushdown(u);
    int ans = INT_MAX;
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid) ans = min(ans, query(u << 1, l, r));
    if (r > mid) ans = min(ans, query(u << 1 | 1, l, r));
    return ans;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    m--;
    build(1, 1, m);
    for (int i = 0; i < n; ++i) {
      cin >> seg[i].l >> seg[i].r >> seg[i].w;
      seg[i].r--;
    }
    sort(seg, seg + n);
    int ans = INT_MAX;
    for (int l = 0, r = 0; r < n; ++r) {
      add(1, seg[r].l, seg[r].r, 1);
      while (l <= r && query(1, 1, m) > 0) {
        ans = min(ans, seg[r].w - seg[l].w);
        add(1, seg[l].l, seg[l].r, -1);
        l++;
      }
    }
    cout << ans;
    return 0;
  }
  