#include <bits/stdc++.h>
using namespace std;
vector<pair<int, pair<int, int> > > v;
int la[4000005], seg[4000005];
void cal(int lv, int st, int ed) {
  int lv2 = lv * 2;
  if (st != ed) {
    la[lv2] += la[lv];
    la[lv2 + 1] += la[lv];
  }
  seg[lv] += la[lv];
  la[lv] = 0;
}
void up(int st, int ed, int lv, int l, int r, int val) {
  cal(lv, st, ed);
  if (st > r || ed < l) {
    return;
  }
  if (st >= l && ed <= r) {
    la[lv] += val;
    cal(lv, st, ed);
    return;
  }
  int lv2 = lv * 2;
  int md = (st + ed) / 2;
  up(st, md, lv2, l, r, val);
  up(md + 1, ed, lv2 + 1, l, r, val);
  seg[lv] = min(seg[lv2], seg[lv2 + 1]);
}
int main() {
  int t, p;
  scanf("%d%d", &t, &p);
  p--;
  for (int i = 0; i < t; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    b--;
    v.push_back({c, {a, b}});
  }
  sort(v.begin(), v.end());
  int mn = 1e7;
  int po = 0;
  for (int i = 0; i < t; i++) {
    up(1, p, 1, v[i].second.first, v[i].second.second, 1);
    while (seg[1] > 0) {
      up(1, p, 1, v[po].second.first, v[po].second.second, -1);
      mn = min(mn, v[i].first - v[po].first);
      po++;
    }
  }
  printf("%d\n", mn);
}

### Prompt

In cpp, your task is to solve 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;
vector<pair<int, pair<int, int> > > v;
int la[4000005], seg[4000005];
void cal(int lv, int st, int ed) {
  int lv2 = lv * 2;
  if (st != ed) {
    la[lv2] += la[lv];
    la[lv2 + 1] += la[lv];
  }
  seg[lv] += la[lv];
  la[lv] = 0;
}
void up(int st, int ed, int lv, int l, int r, int val) {
  cal(lv, st, ed);
  if (st > r || ed < l) {
    return;
  }
  if (st >= l && ed <= r) {
    la[lv] += val;
    cal(lv, st, ed);
    return;
  }
  int lv2 = lv * 2;
  int md = (st + ed) / 2;
  up(st, md, lv2, l, r, val);
  up(md + 1, ed, lv2 + 1, l, r, val);
  seg[lv] = min(seg[lv2], seg[lv2 + 1]);
}
int main() {
  int t, p;
  scanf("%d%d", &t, &p);
  p--;
  for (int i = 0; i < t; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    b--;
    v.push_back({c, {a, b}});
  }
  sort(v.begin(), v.end());
  int mn = 1e7;
  int po = 0;
  for (int i = 0; i < t; i++) {
    up(1, p, 1, v[i].second.first, v[i].second.second, 1);
    while (seg[1] > 0) {
      up(1, p, 1, v[po].second.first, v[po].second.second, -1);
      mn = min(mn, v[i].first - v[po].first);
      po++;
    }
  }
  printf("%d\n", mn);
}

```