#include <bits/stdc++.h>
using namespace std;
const int dim = 300009, dim1 = 4000009, inf = 1e9;
struct seg {
  int l, r, w;
} v[dim];
int aint[dim1], lazy[dim1];
void push(int nod) {
  if (2 * nod + 1 < dim1) {
    lazy[2 * nod] += lazy[nod];
    lazy[2 * nod + 1] += lazy[nod];
  }
  aint[nod] += lazy[nod];
  lazy[nod] = 0;
}
void Update(int nod, int st, int dr, int l, int r, int val) {
  push(nod);
  if (l >= r) return;
  if (st == l && dr == r) {
    lazy[nod] += val;
    push(nod);
    return;
  }
  int mij = (st + dr) / 2;
  Update(2 * nod, st, mij, l, min(mij, r), val);
  Update(2 * nod + 1, mij, dr, max(mij, l), r, val);
  aint[nod] = min(aint[2 * nod], aint[2 * nod + 1]);
}
bool cmp(seg a, seg b) { return a.w < b.w; }
int get() { return aint[1] + lazy[1]; }
signed main() {
  int n, m, ans = inf;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> v[i].l >> v[i].r >> v[i].w;
    v[i].l--;
    v[i].r--;
  }
  m--;
  sort(v, v + n, cmp);
  int j = 0;
  for (int i = 0; i < n; i++) {
    while (j < n && get() == 0) {
      Update(1, 0, m, v[j].l, v[j].r, 1);
      j++;
    }
    if (get() == 0) {
      break;
    }
    ans = min(ans, v[j - 1].w - v[i].w);
    Update(1, 0, m, v[i].l, v[i].r, -1);
  }
  cout << ans;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 dim = 300009, dim1 = 4000009, inf = 1e9;
struct seg {
  int l, r, w;
} v[dim];
int aint[dim1], lazy[dim1];
void push(int nod) {
  if (2 * nod + 1 < dim1) {
    lazy[2 * nod] += lazy[nod];
    lazy[2 * nod + 1] += lazy[nod];
  }
  aint[nod] += lazy[nod];
  lazy[nod] = 0;
}
void Update(int nod, int st, int dr, int l, int r, int val) {
  push(nod);
  if (l >= r) return;
  if (st == l && dr == r) {
    lazy[nod] += val;
    push(nod);
    return;
  }
  int mij = (st + dr) / 2;
  Update(2 * nod, st, mij, l, min(mij, r), val);
  Update(2 * nod + 1, mij, dr, max(mij, l), r, val);
  aint[nod] = min(aint[2 * nod], aint[2 * nod + 1]);
}
bool cmp(seg a, seg b) { return a.w < b.w; }
int get() { return aint[1] + lazy[1]; }
signed main() {
  int n, m, ans = inf;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> v[i].l >> v[i].r >> v[i].w;
    v[i].l--;
    v[i].r--;
  }
  m--;
  sort(v, v + n, cmp);
  int j = 0;
  for (int i = 0; i < n; i++) {
    while (j < n && get() == 0) {
      Update(1, 0, m, v[j].l, v[j].r, 1);
      j++;
    }
    if (get() == 0) {
      break;
    }
    ans = min(ans, v[j - 1].w - v[i].w);
    Update(1, 0, m, v[i].l, v[i].r, -1);
  }
  cout << ans;
}

```