#include <bits/stdc++.h>
using namespace std;
constexpr int N = 3e5, M = 1e6, max_size = 1 << 21;
int t[max_size], inc[max_size];
struct STOperation {
  int mod, l, r;
  int call(int i = 0, int L = 0, int R = M) {
    int dummy = 1e9, &maybe = !mod ? dummy : t[i];
    if (R <= l || r <= L) {
      return maybe + inc[i];
    } else if (l <= L && R <= r) {
      inc[i] += mod;
      return t[i] + inc[i];
    }
    return (maybe = min(call(2 * i + 1, L, (L + R) / 2),
                        call(2 * i + 2, (L + R) / 2, R))) +
           inc[i];
  }
};
pair<int, pair<int, int>> s[N];
int main() {
  cin.tie(0), ios::sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> s[i].second.first >> s[i].second.second >> s[i].first,
        --s[i].second.first, --s[i].second.second;
  }
  sort(s, s + n);
  int l = 0, r = 0, ans = 1e9;
  while (1) {
    if (STOperation{0, 0, m - 1}.call()) {
      ans = min(ans, s[r - 1].first - s[l].first);
      STOperation{-1, s[l].second.first, s[l].second.second}.call();
      ++l;
    } else if (r == n) {
      break;
    } else {
      STOperation{1, s[r].second.first, s[r].second.second}.call();
      ++r;
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 namespace std;
constexpr int N = 3e5, M = 1e6, max_size = 1 << 21;
int t[max_size], inc[max_size];
struct STOperation {
  int mod, l, r;
  int call(int i = 0, int L = 0, int R = M) {
    int dummy = 1e9, &maybe = !mod ? dummy : t[i];
    if (R <= l || r <= L) {
      return maybe + inc[i];
    } else if (l <= L && R <= r) {
      inc[i] += mod;
      return t[i] + inc[i];
    }
    return (maybe = min(call(2 * i + 1, L, (L + R) / 2),
                        call(2 * i + 2, (L + R) / 2, R))) +
           inc[i];
  }
};
pair<int, pair<int, int>> s[N];
int main() {
  cin.tie(0), ios::sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> s[i].second.first >> s[i].second.second >> s[i].first,
        --s[i].second.first, --s[i].second.second;
  }
  sort(s, s + n);
  int l = 0, r = 0, ans = 1e9;
  while (1) {
    if (STOperation{0, 0, m - 1}.call()) {
      ans = min(ans, s[r - 1].first - s[l].first);
      STOperation{-1, s[l].second.first, s[l].second.second}.call();
      ++l;
    } else if (r == n) {
      break;
    } else {
      STOperation{1, s[r].second.first, s[r].second.second}.call();
      ++r;
    }
  }
  cout << ans << '\n';
  return 0;
}

```