#include <bits/stdc++.h>
using namespace std;
int n, m;
struct node {
  int l, r, w;
  bool operator<(const node &p) const { return w < p.w; }
};
node s[300009];
int mn[4000009], lz[4000009];
void dn(int a) {
  int la = a * 2, ra = a * 2 + 1, &v = lz[a];
  if (!v) return;
  lz[la] += v, lz[ra] += v;
  mn[la] += v, mn[ra] += v;
  v = 0;
}
void upd(int a) { mn[a] = min(mn[a * 2], mn[a * 2 + 1]); }
void add(int a, int l, int r, int ll, int rr, int v) {
  if (l > rr || r < ll) return;
  if (l >= ll && r <= rr) {
    lz[a] += v;
    mn[a] += v;
    return;
  }
  dn(a);
  int mid = (l + r) / 2;
  add(a * 2, l, mid, ll, rr, v);
  add(a * 2 + 1, mid + 1, r, ll, rr, v);
  upd(a);
}
int ask(int a, int l, int r, int ll, int rr) {
  if (l > rr || r < ll) return 0x3F3F3F3F;
  ;
  if (l >= ll && r <= rr) return mn[a];
  dn(a);
  int mid = (l + r) / 2;
  return min(ask(a * 2, l, mid, ll, rr), ask(a * 2 + 1, mid + 1, r, ll, rr));
}
int main() {
  scanf("%d%d", &n, &m);
  m--;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d", &s[i].l, &s[i].r, &s[i].w);
    s[i].r--;
  }
  sort(s + 1, s + n + 1);
  int lp = 1, ans = 0x3F3F3F3F;
  ;
  for (int i = 1; i <= n; i++) {
    add(1, 1, m, s[i].l, s[i].r, 1);
    while (lp <= i && mn[1] > 0) {
      ans = min(ans, s[i].w - s[lp].w);
      add(1, 1, m, s[lp].l, s[lp].r, -1);
      lp++;
    }
  }
  printf("%d\n", ans);
  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;
int n, m;
struct node {
  int l, r, w;
  bool operator<(const node &p) const { return w < p.w; }
};
node s[300009];
int mn[4000009], lz[4000009];
void dn(int a) {
  int la = a * 2, ra = a * 2 + 1, &v = lz[a];
  if (!v) return;
  lz[la] += v, lz[ra] += v;
  mn[la] += v, mn[ra] += v;
  v = 0;
}
void upd(int a) { mn[a] = min(mn[a * 2], mn[a * 2 + 1]); }
void add(int a, int l, int r, int ll, int rr, int v) {
  if (l > rr || r < ll) return;
  if (l >= ll && r <= rr) {
    lz[a] += v;
    mn[a] += v;
    return;
  }
  dn(a);
  int mid = (l + r) / 2;
  add(a * 2, l, mid, ll, rr, v);
  add(a * 2 + 1, mid + 1, r, ll, rr, v);
  upd(a);
}
int ask(int a, int l, int r, int ll, int rr) {
  if (l > rr || r < ll) return 0x3F3F3F3F;
  ;
  if (l >= ll && r <= rr) return mn[a];
  dn(a);
  int mid = (l + r) / 2;
  return min(ask(a * 2, l, mid, ll, rr), ask(a * 2 + 1, mid + 1, r, ll, rr));
}
int main() {
  scanf("%d%d", &n, &m);
  m--;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d", &s[i].l, &s[i].r, &s[i].w);
    s[i].r--;
  }
  sort(s + 1, s + n + 1);
  int lp = 1, ans = 0x3F3F3F3F;
  ;
  for (int i = 1; i <= n; i++) {
    add(1, 1, m, s[i].l, s[i].r, 1);
    while (lp <= i && mn[1] > 0) {
      ans = min(ans, s[i].w - s[lp].w);
      add(1, 1, m, s[lp].l, s[lp].r, -1);
      lp++;
    }
  }
  printf("%d\n", ans);
  return 0;
}

```