#include <bits/stdc++.h>
using namespace std;
int n, m;
struct Seg {
  int l;
  int r;
  int w;
};
vector<int> segTree;
vector<int> lazyTree;
void updateSegT(int node, int st, int end, int l, int r, int val) {
  if (l > r) return;
  if (lazyTree[node]) {
    lazyTree[2 * node] += lazyTree[node];
    lazyTree[2 * node + 1] += lazyTree[node];
    segTree[node] += lazyTree[node];
    lazyTree[node] = 0;
  }
  if (l > end || r < st) return;
  if (st >= l && end <= r) {
    segTree[node] += val;
    lazyTree[2 * node] += val;
    lazyTree[2 * node + 1] += val;
  } else {
    int mid = (st + end) / 2;
    updateSegT(2 * node, st, mid, l, r, val);
    updateSegT(2 * node + 1, mid + 1, end, l, r, val);
    segTree[node] = min(segTree[2 * node], segTree[2 * node + 1]);
  }
}
vector<Seg> segs;
int main() {
  cin >> n >> m;
  m--;
  int nm = 1 << (int)ceil(log2(m));
  segTree.resize(nm * 2);
  lazyTree.resize(nm * 2 * 2);
  int l, r, w;
  for (int(i) = (0); (i) < (n); ++(i)) {
    cin >> l >> r >> w;
    segs.push_back({l, r - 1, w});
  }
  sort(segs.begin(), segs.end(),
       [](const Seg& a, const Seg& b) { return (a.w < b.w); });
  updateSegT(1, 1, nm, m + 1, nm, 3000001);
  int ans = INT_MAX;
  for (int p1 = 0, p2 = 0; p1 < n; p1++) {
    updateSegT(1, 1, nm, segs[p1].l, segs[p1].r, 1);
    if (!segTree[1]) continue;
    while (segTree[1]) {
      updateSegT(1, 1, nm, segs[p2].l, segs[p2].r, -1);
      p2++;
    }
    ans = min(ans, segs[p1].w - segs[p2 - 1].w);
  }
  cout << ans;
  return 0;
}

### Prompt

Your task is to create 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;
int n, m;
struct Seg {
  int l;
  int r;
  int w;
};
vector<int> segTree;
vector<int> lazyTree;
void updateSegT(int node, int st, int end, int l, int r, int val) {
  if (l > r) return;
  if (lazyTree[node]) {
    lazyTree[2 * node] += lazyTree[node];
    lazyTree[2 * node + 1] += lazyTree[node];
    segTree[node] += lazyTree[node];
    lazyTree[node] = 0;
  }
  if (l > end || r < st) return;
  if (st >= l && end <= r) {
    segTree[node] += val;
    lazyTree[2 * node] += val;
    lazyTree[2 * node + 1] += val;
  } else {
    int mid = (st + end) / 2;
    updateSegT(2 * node, st, mid, l, r, val);
    updateSegT(2 * node + 1, mid + 1, end, l, r, val);
    segTree[node] = min(segTree[2 * node], segTree[2 * node + 1]);
  }
}
vector<Seg> segs;
int main() {
  cin >> n >> m;
  m--;
  int nm = 1 << (int)ceil(log2(m));
  segTree.resize(nm * 2);
  lazyTree.resize(nm * 2 * 2);
  int l, r, w;
  for (int(i) = (0); (i) < (n); ++(i)) {
    cin >> l >> r >> w;
    segs.push_back({l, r - 1, w});
  }
  sort(segs.begin(), segs.end(),
       [](const Seg& a, const Seg& b) { return (a.w < b.w); });
  updateSegT(1, 1, nm, m + 1, nm, 3000001);
  int ans = INT_MAX;
  for (int p1 = 0, p2 = 0; p1 < n; p1++) {
    updateSegT(1, 1, nm, segs[p1].l, segs[p1].r, 1);
    if (!segTree[1]) continue;
    while (segTree[1]) {
      updateSegT(1, 1, nm, segs[p2].l, segs[p2].r, -1);
      p2++;
    }
    ans = min(ans, segs[p1].w - segs[p2 - 1].w);
  }
  cout << ans;
  return 0;
}

```