#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &out, vector<T> &v) {
  out << "[";
  for (auto k : v) out << k << " ";
  out << "]"
      << "\n";
  return out;
}
template <class T>
ostream &operator<<(ostream &out, set<T> s) {
  out << "{";
  for (auto k : s) out << k << " ";
  out << "}"
      << "\n";
  return out;
}
template <class T, class U>
ostream &operator<<(ostream &out, pair<T, U> p) {
  out << "[ " << p.first << " , " << p.second << " ] ";
  return out;
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &p) {
  in >> p.first >> p.second;
  return in;
}
struct STree {
  vector<int> st, lazy;
  int n;
  STree(int n) : st(4 * n + 5, 0), lazy(4 * n + 5, 0), n(n) {}
  void init(int k, int s, int e, int *a) {
    lazy[k] = 0;
    if (s + 1 == e) {
      st[k] = a[s];
      return;
    }
    int m = (s + e) / 2;
    init(2 * k, s, m, a);
    init(2 * k + 1, m, e, a);
    st[k] = min(st[2 * k], st[2 * k + 1]);
  }
  void push(int k, int s, int e) {
    if (!lazy[k]) return;
    st[k] += lazy[k];
    if (s + 1 < e) {
      lazy[2 * k] += lazy[k];
      lazy[2 * k + 1] += lazy[k];
    }
    lazy[k] = 0;
  }
  void upd(int k, int s, int e, int a, int b, int v) {
    push(k, s, e);
    if (s >= b || e <= a) return;
    if (s >= a && e <= b) {
      lazy[k] += v;
      push(k, s, e);
      return;
    }
    int m = (s + e) / 2;
    upd(2 * k, s, m, a, b, v);
    upd(2 * k + 1, m, e, a, b, v);
    st[k] = min(st[2 * k], st[2 * k + 1]);
  }
  int query(int k, int s, int e, int a, int b) {
    if (s >= b || e <= a) return 1e9;
    push(k, s, e);
    if (s >= a && e <= b) return st[k];
    int m = (s + e) / 2;
    return min(query(2 * k, s, m, a, b), query(2 * k + 1, m, e, a, b));
  }
  void init(int *a) { init(1, 0, n, a); }
  void upd(int a, int b, int v) { upd(1, 0, n, a, b, v); }
  int query(int a, int b) { return query(1, 0, n, a, b); }
};
struct seg {
  int w, l, r;
  void Leer() { cin >> l >> r >> w; }
};
bool operator<(const seg &a, const seg &b) { return a.w < b.w; }
seg in[500000];
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  m++;
  STree st(m);
  for (int i = 0; i < int(n); i++) in[i].Leer();
  sort(in, in + n);
  int i = 0, j = -1;
  int res = 2e6;
  while (i < n) {
    if (st.query(1, m - 1) > 0) {
      res = min(res, in[j].w - in[i].w);
      st.upd(in[i].l, in[i].r, -1);
      i++;
    } else {
      if (j == n - 1) break;
      j++;
      st.upd(in[j].l, in[j].r, 1);
    }
  }
  cout << res << '\n';
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
template <class T>
ostream &operator<<(ostream &out, vector<T> &v) {
  out << "[";
  for (auto k : v) out << k << " ";
  out << "]"
      << "\n";
  return out;
}
template <class T>
ostream &operator<<(ostream &out, set<T> s) {
  out << "{";
  for (auto k : s) out << k << " ";
  out << "}"
      << "\n";
  return out;
}
template <class T, class U>
ostream &operator<<(ostream &out, pair<T, U> p) {
  out << "[ " << p.first << " , " << p.second << " ] ";
  return out;
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &p) {
  in >> p.first >> p.second;
  return in;
}
struct STree {
  vector<int> st, lazy;
  int n;
  STree(int n) : st(4 * n + 5, 0), lazy(4 * n + 5, 0), n(n) {}
  void init(int k, int s, int e, int *a) {
    lazy[k] = 0;
    if (s + 1 == e) {
      st[k] = a[s];
      return;
    }
    int m = (s + e) / 2;
    init(2 * k, s, m, a);
    init(2 * k + 1, m, e, a);
    st[k] = min(st[2 * k], st[2 * k + 1]);
  }
  void push(int k, int s, int e) {
    if (!lazy[k]) return;
    st[k] += lazy[k];
    if (s + 1 < e) {
      lazy[2 * k] += lazy[k];
      lazy[2 * k + 1] += lazy[k];
    }
    lazy[k] = 0;
  }
  void upd(int k, int s, int e, int a, int b, int v) {
    push(k, s, e);
    if (s >= b || e <= a) return;
    if (s >= a && e <= b) {
      lazy[k] += v;
      push(k, s, e);
      return;
    }
    int m = (s + e) / 2;
    upd(2 * k, s, m, a, b, v);
    upd(2 * k + 1, m, e, a, b, v);
    st[k] = min(st[2 * k], st[2 * k + 1]);
  }
  int query(int k, int s, int e, int a, int b) {
    if (s >= b || e <= a) return 1e9;
    push(k, s, e);
    if (s >= a && e <= b) return st[k];
    int m = (s + e) / 2;
    return min(query(2 * k, s, m, a, b), query(2 * k + 1, m, e, a, b));
  }
  void init(int *a) { init(1, 0, n, a); }
  void upd(int a, int b, int v) { upd(1, 0, n, a, b, v); }
  int query(int a, int b) { return query(1, 0, n, a, b); }
};
struct seg {
  int w, l, r;
  void Leer() { cin >> l >> r >> w; }
};
bool operator<(const seg &a, const seg &b) { return a.w < b.w; }
seg in[500000];
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  m++;
  STree st(m);
  for (int i = 0; i < int(n); i++) in[i].Leer();
  sort(in, in + n);
  int i = 0, j = -1;
  int res = 2e6;
  while (i < n) {
    if (st.query(1, m - 1) > 0) {
      res = min(res, in[j].w - in[i].w);
      st.upd(in[i].l, in[i].r, -1);
      i++;
    } else {
      if (j == n - 1) break;
      j++;
      st.upd(in[j].l, in[j].r, 1);
    }
  }
  cout << res << '\n';
  return 0;
}

```