#include <bits/stdc++.h>
using namespace std;
long long int dx[] = {0, 1, 0, -1};
long long int dy[] = {1, 0, -1, 0};
long long int dxx[] = {0, 1, 0, -1, 1, 1, -1, -1};
long long int dyy[] = {1, 0, -1, 0, 1, -1, 1, -1};
const int mxx = 1000006;
int tree[4 * mxx];
void update(int l, int r, int L, int R, int node, int val) {
  if (l > R || r < L)
    return;
  else if (l >= L and r <= R) {
    tree[node] += val;
    return;
  }
  tree[node] -= min(tree[node << 1], tree[(node * 2 + 1)]);
  update(l, (l + r) / 2, L, R, node << 1, val);
  update((l + r) / 2 + 1, r, L, R, (node << 1) + 1, val);
  tree[node] += min(tree[node << 1], tree[(node << 1) + 1]);
}
int query(int l, int r, int L, int R, int node, int mx) {
  if (l > R || r < L)
    return 400000;
  else if (l >= L and r <= R) {
    return tree[node] + mx;
  }
  mx += (tree[node] - min(tree[node << 1], tree[node * 2 + 1]));
  int x = min(query(l, (l + r) / 2, L, R, node * 2, mx),
              query((l + r) / 2 + 1, r, L, R, node * 2 + 1, mx));
  return x;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<pair<int, pair<int, int> > > v;
  for (int i = 1; i <= n; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    v.push_back({c, {a, b}});
  }
  sort(v.begin(), v.end());
  int ok = 0;
  int ans = 1000000;
  int j = 0;
  for (int i = 0; i < n; i++) {
    auto it = v[i];
    update(1, m - 1, it.second.first, it.second.second - 1, 1, 1);
    int x = tree[1];
    if (it.second.second == m) {
      ok = 1;
    }
    if (ok == 0 || x < 1) continue;
    while (query(1, m - 1, v[j].second.first, v[j].second.second - 1, 1, 0) >
           1) {
      update(1, m - 1, v[j].second.first, v[j].second.second - 1, 1, -1);
      j++;
    }
    ans = min(ans, v[i].first - v[j].first);
  }
  printf("%d\n", ans);
}

### Prompt

Create a solution in Cpp for 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;
long long int dx[] = {0, 1, 0, -1};
long long int dy[] = {1, 0, -1, 0};
long long int dxx[] = {0, 1, 0, -1, 1, 1, -1, -1};
long long int dyy[] = {1, 0, -1, 0, 1, -1, 1, -1};
const int mxx = 1000006;
int tree[4 * mxx];
void update(int l, int r, int L, int R, int node, int val) {
  if (l > R || r < L)
    return;
  else if (l >= L and r <= R) {
    tree[node] += val;
    return;
  }
  tree[node] -= min(tree[node << 1], tree[(node * 2 + 1)]);
  update(l, (l + r) / 2, L, R, node << 1, val);
  update((l + r) / 2 + 1, r, L, R, (node << 1) + 1, val);
  tree[node] += min(tree[node << 1], tree[(node << 1) + 1]);
}
int query(int l, int r, int L, int R, int node, int mx) {
  if (l > R || r < L)
    return 400000;
  else if (l >= L and r <= R) {
    return tree[node] + mx;
  }
  mx += (tree[node] - min(tree[node << 1], tree[node * 2 + 1]));
  int x = min(query(l, (l + r) / 2, L, R, node * 2, mx),
              query((l + r) / 2 + 1, r, L, R, node * 2 + 1, mx));
  return x;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<pair<int, pair<int, int> > > v;
  for (int i = 1; i <= n; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    v.push_back({c, {a, b}});
  }
  sort(v.begin(), v.end());
  int ok = 0;
  int ans = 1000000;
  int j = 0;
  for (int i = 0; i < n; i++) {
    auto it = v[i];
    update(1, m - 1, it.second.first, it.second.second - 1, 1, 1);
    int x = tree[1];
    if (it.second.second == m) {
      ok = 1;
    }
    if (ok == 0 || x < 1) continue;
    while (query(1, m - 1, v[j].second.first, v[j].second.second - 1, 1, 0) >
           1) {
      update(1, m - 1, v[j].second.first, v[j].second.second - 1, 1, -1);
      j++;
    }
    ans = min(ans, v[i].first - v[j].first);
  }
  printf("%d\n", ans);
}

```