#include <bits/stdc++.h>
using LL = long long;
class SegTree {
  int n;
  std::vector<int> mx, tag;
  void pushDown(int p) {
    mx[p << 1] += tag[p];
    mx[p << 1 | 1] += tag[p];
    tag[p << 1] += tag[p];
    tag[p << 1 | 1] += tag[p];
    tag[p] = 0;
  }
  void pushUp(int p) { mx[p] = std::min(mx[p << 1], mx[p << 1 | 1]); }
  void modify(int x, int y, int val, int l, int r, int p) {
    if (x <= l && y >= r) {
      tag[p] += val;
      mx[p] += val;
      return;
    }
    pushDown(p);
    int m = (l + r) / 2;
    if (x < m) modify(x, y, val, l, m, p << 1);
    if (m < y) modify(x, y, val, m, r, p << 1 | 1);
    pushUp(p);
  }

 public:
  SegTree(int _n) : n(_n), mx(n * 4), tag(n * 4) {}
  void modify(int x, int y, int val) { modify(--x, y, val, 0, n, 1); }
  int max() { return mx[1]; }
};
void solve() {
  int n, m;
  std::cin >> n >> m;
  SegTree A(m - 1);
  std::vector<std::tuple<int, int, int>> a(n);
  for (auto &[w, l, r] : a) {
    std::cin >> l >> r >> w;
    --r;
  }
  std::sort(a.begin(), a.end());
  int ans = INT_MAX;
  for (int l = 0, r = 0; r < n; ++r) {
    A.modify(std::get<1>(a[r]), std::get<2>(a[r]), 1);
    while (A.max()) {
      ans = std::min(ans, std::get<0>(a[r]) - std::get<0>(a[l]));
      A.modify(std::get<1>(a[l]), std::get<2>(a[l]), -1);
      ++l;
    }
  }
  std::cout << ans << '\n';
}
int main() {
  std::cin.tie(nullptr)->sync_with_stdio(false);
  int cas = 1;
  while (cas--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using LL = long long;
class SegTree {
  int n;
  std::vector<int> mx, tag;
  void pushDown(int p) {
    mx[p << 1] += tag[p];
    mx[p << 1 | 1] += tag[p];
    tag[p << 1] += tag[p];
    tag[p << 1 | 1] += tag[p];
    tag[p] = 0;
  }
  void pushUp(int p) { mx[p] = std::min(mx[p << 1], mx[p << 1 | 1]); }
  void modify(int x, int y, int val, int l, int r, int p) {
    if (x <= l && y >= r) {
      tag[p] += val;
      mx[p] += val;
      return;
    }
    pushDown(p);
    int m = (l + r) / 2;
    if (x < m) modify(x, y, val, l, m, p << 1);
    if (m < y) modify(x, y, val, m, r, p << 1 | 1);
    pushUp(p);
  }

 public:
  SegTree(int _n) : n(_n), mx(n * 4), tag(n * 4) {}
  void modify(int x, int y, int val) { modify(--x, y, val, 0, n, 1); }
  int max() { return mx[1]; }
};
void solve() {
  int n, m;
  std::cin >> n >> m;
  SegTree A(m - 1);
  std::vector<std::tuple<int, int, int>> a(n);
  for (auto &[w, l, r] : a) {
    std::cin >> l >> r >> w;
    --r;
  }
  std::sort(a.begin(), a.end());
  int ans = INT_MAX;
  for (int l = 0, r = 0; r < n; ++r) {
    A.modify(std::get<1>(a[r]), std::get<2>(a[r]), 1);
    while (A.max()) {
      ans = std::min(ans, std::get<0>(a[r]) - std::get<0>(a[l]));
      A.modify(std::get<1>(a[l]), std::get<2>(a[l]), -1);
      ++l;
    }
  }
  std::cout << ans << '\n';
}
int main() {
  std::cin.tie(nullptr)->sync_with_stdio(false);
  int cas = 1;
  while (cas--) {
    solve();
  }
  return 0;
}

```