#include <bits/stdc++.h>
using namespace std;
constexpr int N = 1e6 + 10;
constexpr int MOD = 1e9 + 7;
int tree[8 * N], lazy[8 * N], n, m, ans = 1e9;
vector<pair<int, pair<int, int>>> vec;
void upd(int l, int r, int mid, int idx) {
  int lidx = idx << 1, ridx = lidx | 1;
  if (lazy[idx] == 0) return;
  lazy[lidx] = lazy[idx];
  lazy[ridx] = lazy[idx];
  tree[lidx] = lazy[idx];
  tree[ridx] = lazy[idx];
  lazy[idx] = 0;
}
void update(int b, int e, int l, int r, int val, int idx) {
  if (r <= b || l >= e) return;
  if (b <= l && r <= e) {
    tree[idx] = val;
    lazy[idx] = val;
    return;
  }
  int mid = (l + r) >> 1, lidx = idx << 1, ridx = lidx | 1;
  upd(l, r, mid, idx);
  update(b, e, l, mid, val, lidx);
  update(b, e, mid, r, val, ridx);
  tree[idx] = min(tree[lidx], tree[ridx]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int l, r, w;
    cin >> l >> r >> w;
    l = 2 * l - 1;
    r = 2 * r - 1;
    vec.push_back({w, {l - 1, r}});
  }
  sort(vec.begin(), vec.end());
  int le = -1, ri = 0;
  while (le < n) {
    le++;
    while (tree[1] <= le && ri < n) {
      update(vec[ri].second.first, vec[ri].second.second, 0, 2 * m - 1, ri + 1,
             1);
      ri++;
    }
    if (tree[1] > le) {
      ans = min(ans, vec[ri - 1].first - vec[le].first);
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
constexpr int N = 1e6 + 10;
constexpr int MOD = 1e9 + 7;
int tree[8 * N], lazy[8 * N], n, m, ans = 1e9;
vector<pair<int, pair<int, int>>> vec;
void upd(int l, int r, int mid, int idx) {
  int lidx = idx << 1, ridx = lidx | 1;
  if (lazy[idx] == 0) return;
  lazy[lidx] = lazy[idx];
  lazy[ridx] = lazy[idx];
  tree[lidx] = lazy[idx];
  tree[ridx] = lazy[idx];
  lazy[idx] = 0;
}
void update(int b, int e, int l, int r, int val, int idx) {
  if (r <= b || l >= e) return;
  if (b <= l && r <= e) {
    tree[idx] = val;
    lazy[idx] = val;
    return;
  }
  int mid = (l + r) >> 1, lidx = idx << 1, ridx = lidx | 1;
  upd(l, r, mid, idx);
  update(b, e, l, mid, val, lidx);
  update(b, e, mid, r, val, ridx);
  tree[idx] = min(tree[lidx], tree[ridx]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int l, r, w;
    cin >> l >> r >> w;
    l = 2 * l - 1;
    r = 2 * r - 1;
    vec.push_back({w, {l - 1, r}});
  }
  sort(vec.begin(), vec.end());
  int le = -1, ri = 0;
  while (le < n) {
    le++;
    while (tree[1] <= le && ri < n) {
      update(vec[ri].second.first, vec[ri].second.second, 0, 2 * m - 1, ri + 1,
             1);
      ri++;
    }
    if (tree[1] > le) {
      ans = min(ans, vec[ri - 1].first - vec[le].first);
    }
  }
  cout << ans << '\n';
  return 0;
}

```