#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int inf = 2147483647;
const int maxN = 3e5 + 2;
const int offset = (1 << 21);
int n, m, l[maxN], r[maxN], w[maxN];
int st[offset], lazy[offset];
int ans = inf, sum;
void prop(int x, bool state) {
  if (lazy[x]) st[x] += lazy[x];
  if (!state) {
    lazy[x * 2 + 1] += lazy[x];
    lazy[x * 2 + 2] += lazy[x];
  }
  lazy[x] = 0;
}
void upd(int x, int l, int r, int ql, int qr, int val) {
  prop(x, l == r);
  if (ql <= l && r <= qr) {
    st[x] += val;
    if (l != r) {
      lazy[x * 2 + 1] += val;
      lazy[x * 2 + 2] += val;
    }
    return;
  }
  if (ql > r || qr < l) return;
  int mid = (l + r) >> 1;
  upd(x * 2 + 1, l, mid, ql, qr, val);
  upd(x * 2 + 2, mid + 1, r, ql, qr, val);
  st[x] = min(st[x * 2 + 1], st[x * 2 + 2]);
}
struct Seg {
  int l;
  int r;
  int w;
  inline void load() {
    int a, b, c;
    cin >> a >> b >> c;
    l = a - 1;
    r = b - 2;
    w = c;
  }
} a[maxN];
bool cmp(Seg x, Seg y) { return x.w < y.w; }
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    a[i].load();
  }
  sort(a, a + n, cmp);
  for (int i = 0; i < n; ++i) {
    l[i] = a[i].l;
    r[i] = a[i].r;
    w[i] = a[i].w;
  }
  int p1 = 0, p2 = 0;
  while (p2 < n) {
    if (st[0] == 0) {
      upd(0, 0, m - 2, l[p2], r[p2], 1);
      sum += w[p2];
      p2++;
    } else {
      upd(0, 0, m - 2, l[p1], r[p1], -1);
      ans = min(ans, w[p2 - 1] - w[p1]);
      sum -= w[p1];
      p1++;
    }
  }
  while (p1 < n) {
    if (st[0]) ans = min(ans, w[p2 - 1] - w[p1]);
    upd(0, 0, m - 2, l[p1], r[p1], -1);
    sum -= w[p1];
    p1++;
  }
  cout << ans;
  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 int MOD = 1e9 + 7;
const int inf = 2147483647;
const int maxN = 3e5 + 2;
const int offset = (1 << 21);
int n, m, l[maxN], r[maxN], w[maxN];
int st[offset], lazy[offset];
int ans = inf, sum;
void prop(int x, bool state) {
  if (lazy[x]) st[x] += lazy[x];
  if (!state) {
    lazy[x * 2 + 1] += lazy[x];
    lazy[x * 2 + 2] += lazy[x];
  }
  lazy[x] = 0;
}
void upd(int x, int l, int r, int ql, int qr, int val) {
  prop(x, l == r);
  if (ql <= l && r <= qr) {
    st[x] += val;
    if (l != r) {
      lazy[x * 2 + 1] += val;
      lazy[x * 2 + 2] += val;
    }
    return;
  }
  if (ql > r || qr < l) return;
  int mid = (l + r) >> 1;
  upd(x * 2 + 1, l, mid, ql, qr, val);
  upd(x * 2 + 2, mid + 1, r, ql, qr, val);
  st[x] = min(st[x * 2 + 1], st[x * 2 + 2]);
}
struct Seg {
  int l;
  int r;
  int w;
  inline void load() {
    int a, b, c;
    cin >> a >> b >> c;
    l = a - 1;
    r = b - 2;
    w = c;
  }
} a[maxN];
bool cmp(Seg x, Seg y) { return x.w < y.w; }
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    a[i].load();
  }
  sort(a, a + n, cmp);
  for (int i = 0; i < n; ++i) {
    l[i] = a[i].l;
    r[i] = a[i].r;
    w[i] = a[i].w;
  }
  int p1 = 0, p2 = 0;
  while (p2 < n) {
    if (st[0] == 0) {
      upd(0, 0, m - 2, l[p2], r[p2], 1);
      sum += w[p2];
      p2++;
    } else {
      upd(0, 0, m - 2, l[p1], r[p1], -1);
      ans = min(ans, w[p2 - 1] - w[p1]);
      sum -= w[p1];
      p1++;
    }
  }
  while (p1 < n) {
    if (st[0]) ans = min(ans, w[p2 - 1] - w[p1]);
    upd(0, 0, m - 2, l[p1], r[p1], -1);
    sum -= w[p1];
    p1++;
  }
  cout << ans;
  return 0;
}

```