#include <bits/stdc++.h>
using namespace std;
vector<pair<pair<int, int>, int> > a;
int n, m;
int t[4 * 1001000];
int add[4 * 1001000];
void push(int v, int tl, int tr) {
  if (add[v] == 0) return;
  t[v] += add[v];
  if (tl == tr) {
    add[v] = 0;
    return;
  }
  add[2 * v] += add[v];
  add[2 * v + 1] += add[v];
  add[v] = 0;
}
void update(int v, int tl, int tr, int l, int r, int val) {
  push(v, tl, tr);
  if (tl > r || tr < l) return;
  if (l <= tl && tr <= r) {
    add[v] += val;
    return;
  }
  int mid = (tl + tr) >> 1;
  update(2 * v, tl, mid, l, r, val);
  update(2 * v + 1, mid + 1, tr, l, r, val);
  t[v] = min(t[2 * v] + add[2 * v], t[2 * v + 1] + add[2 * v + 1]);
}
int get(int v, int tl, int tr, int l, int r) {
  push(v, tl, tr);
  if (l > tr || r < tl) return (int)1e9;
  if (l <= tl && tr <= r) {
    return t[v] + add[v];
  }
  int mid = (tl + tr) >> 1;
  int ans1 = get(2 * v, tl, mid, l, r);
  int ans2 = get(2 * v + 1, mid + 1, tr, l, r);
  return min(ans1, ans2);
}
void print(int v, int tl, int tr, int tab = 0) {
  for (int i = 0; i < tab; i++) printf("  ");
  printf("(%d, %d) - [%d, %d] v=%d\n", t[v], add[v], tl, tr, v);
  if (tl == tr) return;
  int mid = (tl + tr) >> 1;
  print(2 * v, tl, mid, tab + 1);
  print(2 * v + 1, mid + 1, tr, tab + 1);
}
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.second < b.second;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int l, r, x;
    cin >> l >> r >> x;
    a.push_back({{l, r}, x});
  }
  sort(a.begin(), a.end(), cmp);
  int r = -1;
  int ans = (int)1e9;
  for (int l = 0; l < n; l++) {
    while (r + 1 < n && get(1, 0, m, 1, m - 1) == 0) {
      update(1, 0, m, a[r + 1].first.first, a[r + 1].first.second - 1, 1);
      r++;
    }
    if (get(1, 0, m, 1, m - 1) == 0) break;
    ans = min(ans, a[r].second - a[l].second);
    update(1, 0, m, a[l].first.first, a[l].first.second - 1, -1);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<pair<pair<int, int>, int> > a;
int n, m;
int t[4 * 1001000];
int add[4 * 1001000];
void push(int v, int tl, int tr) {
  if (add[v] == 0) return;
  t[v] += add[v];
  if (tl == tr) {
    add[v] = 0;
    return;
  }
  add[2 * v] += add[v];
  add[2 * v + 1] += add[v];
  add[v] = 0;
}
void update(int v, int tl, int tr, int l, int r, int val) {
  push(v, tl, tr);
  if (tl > r || tr < l) return;
  if (l <= tl && tr <= r) {
    add[v] += val;
    return;
  }
  int mid = (tl + tr) >> 1;
  update(2 * v, tl, mid, l, r, val);
  update(2 * v + 1, mid + 1, tr, l, r, val);
  t[v] = min(t[2 * v] + add[2 * v], t[2 * v + 1] + add[2 * v + 1]);
}
int get(int v, int tl, int tr, int l, int r) {
  push(v, tl, tr);
  if (l > tr || r < tl) return (int)1e9;
  if (l <= tl && tr <= r) {
    return t[v] + add[v];
  }
  int mid = (tl + tr) >> 1;
  int ans1 = get(2 * v, tl, mid, l, r);
  int ans2 = get(2 * v + 1, mid + 1, tr, l, r);
  return min(ans1, ans2);
}
void print(int v, int tl, int tr, int tab = 0) {
  for (int i = 0; i < tab; i++) printf("  ");
  printf("(%d, %d) - [%d, %d] v=%d\n", t[v], add[v], tl, tr, v);
  if (tl == tr) return;
  int mid = (tl + tr) >> 1;
  print(2 * v, tl, mid, tab + 1);
  print(2 * v + 1, mid + 1, tr, tab + 1);
}
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.second < b.second;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int l, r, x;
    cin >> l >> r >> x;
    a.push_back({{l, r}, x});
  }
  sort(a.begin(), a.end(), cmp);
  int r = -1;
  int ans = (int)1e9;
  for (int l = 0; l < n; l++) {
    while (r + 1 < n && get(1, 0, m, 1, m - 1) == 0) {
      update(1, 0, m, a[r + 1].first.first, a[r + 1].first.second - 1, 1);
      r++;
    }
    if (get(1, 0, m, 1, m - 1) == 0) break;
    ans = min(ans, a[r].second - a[l].second);
    update(1, 0, m, a[l].first.first, a[l].first.second - 1, -1);
  }
  cout << ans << endl;
  return 0;
}

```