#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 7;
int n, m;
typedef struct Edge {
  int l, r, w;
} Edge;
bool cmp(Edge a, Edge b) { return a.w < b.w; }
Edge eg[maxn];
int tot = 0;
int uni[maxn * 2];
int t[maxn << 3];
int lazy[maxn << 3];
void pushDown(int k);
void update(int k, int l, int r, int x, int y, int v);
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> eg[i].l >> eg[i].r >> eg[i].w;
  }
  m--;
  sort(eg + 1, eg + 1 + n, cmp);
  int p, q;
  p = q = 1;
  int ans = 1e9;
  update(1, 1, m, eg[q].l, eg[q].r - 1, 1);
  while (p <= q && q <= n) {
    while (t[1] >= 1 && p <= q) {
      ans = min(ans, eg[q].w - eg[p].w);
      update(1, 1, m, eg[p].l, eg[p].r - 1, -1);
      p++;
    }
    q++;
    if (q > n) break;
    update(1, 1, m, eg[q].l, eg[q].r - 1, 1);
  }
  if (ans == 1e9)
    cout << -1;
  else
    cout << ans;
}
void update(int k, int l, int r, int x, int y, int v) {
  if (x <= l && r <= y) {
    t[k] += v;
    lazy[k] += v;
  } else {
    pushDown(k);
    int mid = (l + r) / 2;
    if (x <= mid) update(k * 2, l, mid, x, y, v);
    if (y > mid) update(k * 2 + 1, mid + 1, r, x, y, v);
    t[k] = min(t[k * 2 + 1], t[k * 2]);
  }
}
void pushDown(int k) {
  if (lazy[k]) {
    lazy[k << 1] += lazy[k];
    lazy[k << 1 | 1] += lazy[k];
    t[k << 1] += lazy[k];
    t[k << 1 | 1] += lazy[k];
    lazy[k] = 0;
  }
}

### Prompt

Please provide a cpp coded solution 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;
const int maxn = 5e5 + 7;
int n, m;
typedef struct Edge {
  int l, r, w;
} Edge;
bool cmp(Edge a, Edge b) { return a.w < b.w; }
Edge eg[maxn];
int tot = 0;
int uni[maxn * 2];
int t[maxn << 3];
int lazy[maxn << 3];
void pushDown(int k);
void update(int k, int l, int r, int x, int y, int v);
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> eg[i].l >> eg[i].r >> eg[i].w;
  }
  m--;
  sort(eg + 1, eg + 1 + n, cmp);
  int p, q;
  p = q = 1;
  int ans = 1e9;
  update(1, 1, m, eg[q].l, eg[q].r - 1, 1);
  while (p <= q && q <= n) {
    while (t[1] >= 1 && p <= q) {
      ans = min(ans, eg[q].w - eg[p].w);
      update(1, 1, m, eg[p].l, eg[p].r - 1, -1);
      p++;
    }
    q++;
    if (q > n) break;
    update(1, 1, m, eg[q].l, eg[q].r - 1, 1);
  }
  if (ans == 1e9)
    cout << -1;
  else
    cout << ans;
}
void update(int k, int l, int r, int x, int y, int v) {
  if (x <= l && r <= y) {
    t[k] += v;
    lazy[k] += v;
  } else {
    pushDown(k);
    int mid = (l + r) / 2;
    if (x <= mid) update(k * 2, l, mid, x, y, v);
    if (y > mid) update(k * 2 + 1, mid + 1, r, x, y, v);
    t[k] = min(t[k * 2 + 1], t[k * 2]);
  }
}
void pushDown(int k) {
  if (lazy[k]) {
    lazy[k << 1] += lazy[k];
    lazy[k << 1 | 1] += lazy[k];
    t[k << 1] += lazy[k];
    t[k << 1 | 1] += lazy[k];
    lazy[k] = 0;
  }
}

```