#include <bits/stdc++.h>
using namespace std;
const long long maxm = 2e6 + 5;
struct Node {
  long long l, r, w;
} e[maxm];
bool cmp(Node a, Node b) { return a.w < b.w; }
struct Tree {
  long long mi[maxm << 2];
  long long laz[maxm << 2];
  void pp(long long node) { mi[node] = min(mi[node * 2], mi[node * 2 + 1]); }
  void pd(long long node) {
    if (laz[node]) {
      mi[node * 2] += laz[node];
      mi[node * 2 + 1] += laz[node];
      laz[node * 2] += laz[node];
      laz[node * 2 + 1] += laz[node];
      laz[node] = 0;
    }
  }
  void update(long long st, long long ed, long long val, long long l,
              long long r, long long node) {
    if (st <= l && ed >= r) {
      laz[node] += val;
      mi[node] += val;
      return;
    }
    pd(node);
    long long mid = (l + r) / 2;
    if (st <= mid) update(st, ed, val, l, mid, node * 2);
    if (ed > mid) update(st, ed, val, mid + 1, r, node * 2 + 1);
    pp(node);
  }
} T;
long long n, m;
void solve() {
  cin >> n >> m;
  m--;
  for (long long i = 1; i <= n; i++) {
    long long l, r, w;
    cin >> l >> r >> w;
    r--;
    e[i] = {l, r, w};
  }
  sort(e + 1, e + 1 + n, cmp);
  long long ans = 1e18;
  long long l = 1;
  for (long long i = 1; i <= n; i++) {
    T.update(e[i].l, e[i].r, 1, 1, m, 1);
    while (T.mi[1] > 0) {
      ans = min(ans, e[i].w - e[l].w);
      T.update(e[l].l, e[l].r, -1, 1, m, 1);
      l++;
    }
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}

### Prompt

Your challenge is to write 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 long long maxm = 2e6 + 5;
struct Node {
  long long l, r, w;
} e[maxm];
bool cmp(Node a, Node b) { return a.w < b.w; }
struct Tree {
  long long mi[maxm << 2];
  long long laz[maxm << 2];
  void pp(long long node) { mi[node] = min(mi[node * 2], mi[node * 2 + 1]); }
  void pd(long long node) {
    if (laz[node]) {
      mi[node * 2] += laz[node];
      mi[node * 2 + 1] += laz[node];
      laz[node * 2] += laz[node];
      laz[node * 2 + 1] += laz[node];
      laz[node] = 0;
    }
  }
  void update(long long st, long long ed, long long val, long long l,
              long long r, long long node) {
    if (st <= l && ed >= r) {
      laz[node] += val;
      mi[node] += val;
      return;
    }
    pd(node);
    long long mid = (l + r) / 2;
    if (st <= mid) update(st, ed, val, l, mid, node * 2);
    if (ed > mid) update(st, ed, val, mid + 1, r, node * 2 + 1);
    pp(node);
  }
} T;
long long n, m;
void solve() {
  cin >> n >> m;
  m--;
  for (long long i = 1; i <= n; i++) {
    long long l, r, w;
    cin >> l >> r >> w;
    r--;
    e[i] = {l, r, w};
  }
  sort(e + 1, e + 1 + n, cmp);
  long long ans = 1e18;
  long long l = 1;
  for (long long i = 1; i <= n; i++) {
    T.update(e[i].l, e[i].r, 1, 1, m, 1);
    while (T.mi[1] > 0) {
      ans = min(ans, e[i].w - e[l].w);
      T.update(e[l].l, e[l].r, -1, 1, m, 1);
      l++;
    }
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}

```