#include <bits/stdc++.h>
using namespace std;
mt19937 rng(std::chrono::steady_clock::now().time_since_epoch().count());
template <typename A, typename B>
ostream& operator<<(ostream& cout, pair<A, B> const& p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream& operator<<(ostream& cout, vector<A> const& v) {
  cout << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
struct LazySeg {
  std::vector<T> tree, lazy;
  int n;
  LazySeg() {}
  LazySeg(int _n) { init(_n); }
  LazySeg(vector<T>& arr) {
    init(arr.size());
    init(arr);
  }
  void init(int _n) {
    n = _n;
    tree.assign(4 * n, 0);
    lazy.assign(4 * n, 0);
  }
  void init(vector<T>& arr, int i = 0, int tl = 0, int tr = -1) {
    if (tr == -1) tr = n - 1;
    if (tl == tr) {
      tree[i] = arr[tl];
      return;
    }
    int mid = (tl + tr) / 2;
    init(arr, 2 * i + 1, tl, mid);
    init(arr, 2 * i + 2, mid + 1, tr);
    pull(i);
  }
  const T ID = 1000000007;
  void push(int i, int tl, int tr) {
    tree[i] += lazy[i];
    if (tl != tr) {
      lazy[2 * i + 1] += lazy[i];
      lazy[2 * i + 2] += lazy[i];
    }
    lazy[i] = 0;
  }
  T comb(T a, T b) { return min(a, b); }
  void pull(int i) { tree[i] = comb(tree[2 * i + 1], tree[2 * i + 2]); }
  void upd(int ql, int qr, T v, int i = 0, int tl = 0, int tr = -1) {
    if (tr == -1) tr = n - 1;
    push(i, tl, tr);
    if (qr < tl || tr < ql) return;
    if (ql <= tl && tr <= qr) {
      lazy[i] = v;
      push(i, tl, tr);
      return;
    }
    int mid = (tl + tr) / 2;
    upd(ql, qr, v, 2 * i + 1, tl, mid);
    upd(ql, qr, v, 2 * i + 2, mid + 1, tr);
    pull(i);
  }
  T query(int ql, int qr, int i = 0, int tl = 0, int tr = -1) {
    if (tr == -1) tr = n - 1;
    push(i, tl, tr);
    if (ql > tr || tl > qr) return ID;
    if (ql <= tl && tr <= qr) return tree[i];
    int mid = (tl + tr) / 2;
    return comb(query(ql, qr, 2 * i + 1, tl, mid),
                query(ql, qr, 2 * i + 2, mid + 1, tr));
  }
};
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, pair<int, int>>> v(n);
  for (int i = 0; i < (n); i++) {
    cin >> v[i].second.first >> v[i].second.second >> v[i].first;
  }
  sort(v.begin(), v.end());
  LazySeg<int> st(m);
  int r = 0;
  int best = 1000000007;
  for (int i = 0; i < (n); i++) {
    while (r < n && st.query(0, m - 2) == 0) {
      st.upd(v[r].second.first - 1, v[r].second.second - 2, 1);
      r++;
    }
    if (st.query(0, m - 2)) {
      ckmin(best, v[r - 1].first - v[i].first);
    }
    st.upd(v[i].second.first - 1, v[i].second.second - 2, -1);
  }
  cout << best << "\n";
}

### Prompt

Please formulate 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(std::chrono::steady_clock::now().time_since_epoch().count());
template <typename A, typename B>
ostream& operator<<(ostream& cout, pair<A, B> const& p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream& operator<<(ostream& cout, vector<A> const& v) {
  cout << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
struct LazySeg {
  std::vector<T> tree, lazy;
  int n;
  LazySeg() {}
  LazySeg(int _n) { init(_n); }
  LazySeg(vector<T>& arr) {
    init(arr.size());
    init(arr);
  }
  void init(int _n) {
    n = _n;
    tree.assign(4 * n, 0);
    lazy.assign(4 * n, 0);
  }
  void init(vector<T>& arr, int i = 0, int tl = 0, int tr = -1) {
    if (tr == -1) tr = n - 1;
    if (tl == tr) {
      tree[i] = arr[tl];
      return;
    }
    int mid = (tl + tr) / 2;
    init(arr, 2 * i + 1, tl, mid);
    init(arr, 2 * i + 2, mid + 1, tr);
    pull(i);
  }
  const T ID = 1000000007;
  void push(int i, int tl, int tr) {
    tree[i] += lazy[i];
    if (tl != tr) {
      lazy[2 * i + 1] += lazy[i];
      lazy[2 * i + 2] += lazy[i];
    }
    lazy[i] = 0;
  }
  T comb(T a, T b) { return min(a, b); }
  void pull(int i) { tree[i] = comb(tree[2 * i + 1], tree[2 * i + 2]); }
  void upd(int ql, int qr, T v, int i = 0, int tl = 0, int tr = -1) {
    if (tr == -1) tr = n - 1;
    push(i, tl, tr);
    if (qr < tl || tr < ql) return;
    if (ql <= tl && tr <= qr) {
      lazy[i] = v;
      push(i, tl, tr);
      return;
    }
    int mid = (tl + tr) / 2;
    upd(ql, qr, v, 2 * i + 1, tl, mid);
    upd(ql, qr, v, 2 * i + 2, mid + 1, tr);
    pull(i);
  }
  T query(int ql, int qr, int i = 0, int tl = 0, int tr = -1) {
    if (tr == -1) tr = n - 1;
    push(i, tl, tr);
    if (ql > tr || tl > qr) return ID;
    if (ql <= tl && tr <= qr) return tree[i];
    int mid = (tl + tr) / 2;
    return comb(query(ql, qr, 2 * i + 1, tl, mid),
                query(ql, qr, 2 * i + 2, mid + 1, tr));
  }
};
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, pair<int, int>>> v(n);
  for (int i = 0; i < (n); i++) {
    cin >> v[i].second.first >> v[i].second.second >> v[i].first;
  }
  sort(v.begin(), v.end());
  LazySeg<int> st(m);
  int r = 0;
  int best = 1000000007;
  for (int i = 0; i < (n); i++) {
    while (r < n && st.query(0, m - 2) == 0) {
      st.upd(v[r].second.first - 1, v[r].second.second - 2, 1);
      r++;
    }
    if (st.query(0, m - 2)) {
      ckmin(best, v[r - 1].first - v[i].first);
    }
    st.upd(v[i].second.first - 1, v[i].second.second - 2, -1);
  }
  cout << best << "\n";
}

```