#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(time(0));
long long mod = 1000000007;
long long inf = 1000000000000000;
const long long MAXN = 2e5 + 10;
struct SegmentTree {
  long long R;
  vector<long long> mns, t;
  void update(long long x, long long xl, long long xr, long long ql,
              long long qr, long long val) {
    if (xl >= qr || ql >= xr) return;
    if (ql <= xl && xr <= qr) {
      t[x] += val;
      mns[x] += val;
      return;
    }
    long long mid = (xl + xr) >> 1;
    update(x * 2, xl, mid, ql, qr, val);
    update(x * 2 + 1, mid, xr, ql, qr, val);
    mns[x] = t[x] + min(mns[x * 2], mns[x * 2 + 1]);
  }
  void upd(tuple<long long, long long, long long> segment, long long val) {
    update(1, 0, R, get<1>(segment) - 1, get<2>(segment) - 1, val);
  }
  SegmentTree(long long m) {
    R = 1;
    while (R < m) R *= 2;
    mns.assign(2 * R, 0);
    t.assign(2 * R, 0);
    update(1, 0, R, m - 1, R, 1);
  }
};
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<tuple<long long, long long, long long>> a(n);
  for (long long i = 0; i < n; ++i) {
    long long l, r, w;
    cin >> l >> r >> w;
    a[i] = {w, l, r};
  }
  sort(a.begin(), a.end());
  SegmentTree sg(m);
  long long r = 0, ans = inf;
  for (long long l = 0; l < n; ++l) {
    while (r < n && sg.mns[1] == 0) sg.upd(a[r++], 1);
    if (sg.mns[1] != 0) ans = min(ans, get<0>(a[r - 1]) - get<0>(a[l]));
    sg.upd(a[l], -1);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  solve();
  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;
mt19937 rnd(time(0));
long long mod = 1000000007;
long long inf = 1000000000000000;
const long long MAXN = 2e5 + 10;
struct SegmentTree {
  long long R;
  vector<long long> mns, t;
  void update(long long x, long long xl, long long xr, long long ql,
              long long qr, long long val) {
    if (xl >= qr || ql >= xr) return;
    if (ql <= xl && xr <= qr) {
      t[x] += val;
      mns[x] += val;
      return;
    }
    long long mid = (xl + xr) >> 1;
    update(x * 2, xl, mid, ql, qr, val);
    update(x * 2 + 1, mid, xr, ql, qr, val);
    mns[x] = t[x] + min(mns[x * 2], mns[x * 2 + 1]);
  }
  void upd(tuple<long long, long long, long long> segment, long long val) {
    update(1, 0, R, get<1>(segment) - 1, get<2>(segment) - 1, val);
  }
  SegmentTree(long long m) {
    R = 1;
    while (R < m) R *= 2;
    mns.assign(2 * R, 0);
    t.assign(2 * R, 0);
    update(1, 0, R, m - 1, R, 1);
  }
};
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<tuple<long long, long long, long long>> a(n);
  for (long long i = 0; i < n; ++i) {
    long long l, r, w;
    cin >> l >> r >> w;
    a[i] = {w, l, r};
  }
  sort(a.begin(), a.end());
  SegmentTree sg(m);
  long long r = 0, ans = inf;
  for (long long l = 0; l < n; ++l) {
    while (r < n && sg.mns[1] == 0) sg.upd(a[r++], 1);
    if (sg.mns[1] != 0) ans = min(ans, get<0>(a[r - 1]) - get<0>(a[l]));
    sg.upd(a[l], -1);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  solve();
  return 0;
}

```