#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
int seg[maxn << 2], laz[maxn << 2], ord[maxn], le[maxn], ri[maxn], we[maxn];
int n, m;
void add(int val, int st, int en, int l = 0, int r = m, int id = 1) {
  if (st >= r or en <= l) return;
  if (st <= l and r <= en) {
    seg[id] += val;
    laz[id] += val;
    return;
  }
  int mid = (l + r) >> 1;
  add(val, st, en, l, mid, id << 1 | 0);
  add(val, st, en, mid, r, id << 1 | 1);
  seg[id] = min(seg[id << 1 | 0], seg[id << 1 | 1]) + laz[id];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  m--;
  for (int i = 0; i < n; i++) {
    cin >> le[i] >> ri[i] >> we[i];
    le[i]--, ri[i]--;
  }
  iota(ord, ord + n, 0);
  sort(ord, ord + n, [](int i, int j) -> bool { return we[i] < we[j]; });
  int ans = 1e6;
  int ptr = 0;
  for (int i = 0; i < n; i++) {
    add(+1, le[ord[i]], ri[ord[i]]);
    while (seg[1] > 0 and ptr <= i) {
      add(-1, le[ord[ptr]], ri[ord[ptr]]);
      ptr++;
    }
    if (ptr - 1 >= 0) (ans) = min((ans), (we[ord[i]] - we[ord[ptr - 1]]));
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Generate a Cpp solution to 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 namespace std;
const int maxn = 1e6 + 10;
int seg[maxn << 2], laz[maxn << 2], ord[maxn], le[maxn], ri[maxn], we[maxn];
int n, m;
void add(int val, int st, int en, int l = 0, int r = m, int id = 1) {
  if (st >= r or en <= l) return;
  if (st <= l and r <= en) {
    seg[id] += val;
    laz[id] += val;
    return;
  }
  int mid = (l + r) >> 1;
  add(val, st, en, l, mid, id << 1 | 0);
  add(val, st, en, mid, r, id << 1 | 1);
  seg[id] = min(seg[id << 1 | 0], seg[id << 1 | 1]) + laz[id];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  m--;
  for (int i = 0; i < n; i++) {
    cin >> le[i] >> ri[i] >> we[i];
    le[i]--, ri[i]--;
  }
  iota(ord, ord + n, 0);
  sort(ord, ord + n, [](int i, int j) -> bool { return we[i] < we[j]; });
  int ans = 1e6;
  int ptr = 0;
  for (int i = 0; i < n; i++) {
    add(+1, le[ord[i]], ri[ord[i]]);
    while (seg[1] > 0 and ptr <= i) {
      add(-1, le[ord[ptr]], ri[ord[ptr]]);
      ptr++;
    }
    if (ptr - 1 >= 0) (ans) = min((ans), (we[ord[i]] - we[ord[ptr - 1]]));
  }
  cout << ans << '\n';
  return 0;
}

```