#include <bits/stdc++.h>
using namespace std;
long long inf = 4e18, mod = 1e9 + 7;
int segt[8000005], lazy[8000005];
void push(int v, int l, int r) {
  segt[v] += lazy[v];
  if (l != r) {
    lazy[2 * v] += lazy[v];
    lazy[2 * v + 1] += lazy[v];
  }
  lazy[v] = 0;
}
void update(int v, int l, int r, int qs, int qe, int x) {
  push(v, l, r);
  if (r < qs || l > qe || qs > qe) return;
  if (qs <= l && r <= qe) {
    lazy[v] += x;
    push(v, l, r);
    return;
  }
  int m = (l + r) / 2;
  update(2 * v, l, m, qs, qe, x);
  update(2 * v + 1, m + 1, r, qs, qe, x);
  segt[v] = min(segt[2 * v], segt[2 * v + 1]);
}
int query(int v, int l, int r, int qs, int qe) {
  push(v, l, r);
  if (r < qs || l > qe || qs > qe) return mod;
  if (qs <= l && r <= qe) return segt[v];
  int m = (l + r) / 2;
  int v1 = query(2 * v, l, m, qs, qe);
  int v2 = query(2 * v + 1, m + 1, r, qs, qe);
  return min(v1, v2);
}
void solve() {
  int n, m;
  cin >> n >> m;
  m *= 2;
  vector<vector<int>> v;
  for (int i = 1; i <= n; i++) {
    int l, r, w;
    cin >> l >> r >> w;
    l *= 2, r *= 2;
    v.push_back({w, l, r});
  }
  sort(v.begin(), v.end());
  int r = 0, res = mod;
  for (int l = 1; l <= n; l++) {
    while (r < n && query(1, 1, m, 2, m) == 0) {
      r++;
      int lf = v[r - 1][1], rf = v[r - 1][2];
      update(1, 1, m, lf, rf, 1);
    }
    if (query(1, 1, m, 2, m) > 0) {
      res = min(res, v[r - 1][0] - v[l - 1][0]);
    }
    int lf = v[l - 1][1], rf = v[l - 1][2];
    update(1, 1, m, lf, rf, -1);
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
}

### 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;
long long inf = 4e18, mod = 1e9 + 7;
int segt[8000005], lazy[8000005];
void push(int v, int l, int r) {
  segt[v] += lazy[v];
  if (l != r) {
    lazy[2 * v] += lazy[v];
    lazy[2 * v + 1] += lazy[v];
  }
  lazy[v] = 0;
}
void update(int v, int l, int r, int qs, int qe, int x) {
  push(v, l, r);
  if (r < qs || l > qe || qs > qe) return;
  if (qs <= l && r <= qe) {
    lazy[v] += x;
    push(v, l, r);
    return;
  }
  int m = (l + r) / 2;
  update(2 * v, l, m, qs, qe, x);
  update(2 * v + 1, m + 1, r, qs, qe, x);
  segt[v] = min(segt[2 * v], segt[2 * v + 1]);
}
int query(int v, int l, int r, int qs, int qe) {
  push(v, l, r);
  if (r < qs || l > qe || qs > qe) return mod;
  if (qs <= l && r <= qe) return segt[v];
  int m = (l + r) / 2;
  int v1 = query(2 * v, l, m, qs, qe);
  int v2 = query(2 * v + 1, m + 1, r, qs, qe);
  return min(v1, v2);
}
void solve() {
  int n, m;
  cin >> n >> m;
  m *= 2;
  vector<vector<int>> v;
  for (int i = 1; i <= n; i++) {
    int l, r, w;
    cin >> l >> r >> w;
    l *= 2, r *= 2;
    v.push_back({w, l, r});
  }
  sort(v.begin(), v.end());
  int r = 0, res = mod;
  for (int l = 1; l <= n; l++) {
    while (r < n && query(1, 1, m, 2, m) == 0) {
      r++;
      int lf = v[r - 1][1], rf = v[r - 1][2];
      update(1, 1, m, lf, rf, 1);
    }
    if (query(1, 1, m, 2, m) > 0) {
      res = min(res, v[r - 1][0] - v[l - 1][0]);
    }
    int lf = v[l - 1][1], rf = v[l - 1][2];
    update(1, 1, m, lf, rf, -1);
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
}

```