#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int P = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
int n, m;
struct Line {
  int l, r, w;
  bool operator<(const Line &t) const { return w < t.w; }
} line[N];
struct Node {
  int l, r;
  int val, lazy;
} tr[N * 4];
void pushup(int u) { tr[u].val = min(tr[u << 1].val, tr[u << 1 | 1].val); }
void pushdown(int u) {
  if (tr[u].lazy != -1) {
    tr[u << 1].val = tr[u << 1 | 1].val = tr[u].lazy;
    tr[u << 1].lazy = tr[u << 1 | 1].lazy = tr[u].lazy;
    tr[u].lazy = -1;
  }
}
void build(int u, int l, int r) {
  tr[u] = {l, r, 0, -1};
  if (l == r)
    return;
  else {
    int mid = l + r >> 1;
    build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
  }
}
void modify(int u, int l, int r, int d) {
  if (tr[u].l >= l && tr[u].r <= r)
    tr[u].val = tr[u].lazy = d;
  else {
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid) modify(u << 1, l, r, d);
    if (r > mid) modify(u << 1 | 1, l, r, d);
    pushup(u);
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> line[i].l >> line[i].r >> line[i].w;
  build(1, 1, m - 1);
  sort(line + 1, line + n + 1);
  int res = INF;
  for (int i = 1; i <= n; ++i) {
    modify(1, line[i].l, line[i].r - 1, line[i].w);
    int t = tr[1].val;
    if (t) res = min(res, line[i].w - t);
  }
  cout << res << endl;
  return 0;
}

### Prompt

Your task is to create 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int P = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
int n, m;
struct Line {
  int l, r, w;
  bool operator<(const Line &t) const { return w < t.w; }
} line[N];
struct Node {
  int l, r;
  int val, lazy;
} tr[N * 4];
void pushup(int u) { tr[u].val = min(tr[u << 1].val, tr[u << 1 | 1].val); }
void pushdown(int u) {
  if (tr[u].lazy != -1) {
    tr[u << 1].val = tr[u << 1 | 1].val = tr[u].lazy;
    tr[u << 1].lazy = tr[u << 1 | 1].lazy = tr[u].lazy;
    tr[u].lazy = -1;
  }
}
void build(int u, int l, int r) {
  tr[u] = {l, r, 0, -1};
  if (l == r)
    return;
  else {
    int mid = l + r >> 1;
    build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
  }
}
void modify(int u, int l, int r, int d) {
  if (tr[u].l >= l && tr[u].r <= r)
    tr[u].val = tr[u].lazy = d;
  else {
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid) modify(u << 1, l, r, d);
    if (r > mid) modify(u << 1 | 1, l, r, d);
    pushup(u);
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> line[i].l >> line[i].r >> line[i].w;
  build(1, 1, m - 1);
  sort(line + 1, line + n + 1);
  int res = INF;
  for (int i = 1; i <= n; ++i) {
    modify(1, line[i].l, line[i].r - 1, line[i].w);
    int t = tr[1].val;
    if (t) res = min(res, line[i].w - t);
  }
  cout << res << endl;
  return 0;
}

```