#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";
}

### Prompt

Please provide a Cpp coded solution 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>
#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";
}

```