#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10, M = 1e6 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int P = 131;
int n, m;
int sum[M * 4], lz[M * 4];
struct node {
  int l, r, w;
  bool operator<(const node &t) const { return w < t.w; }
} seg[N];
void pushdown(int rt) {
  int ls = rt << 1, rs = rt << 1 | 1;
  if (lz[rt]) {
    sum[ls] += lz[rt];
    lz[ls] += lz[rt];
    sum[rs] += lz[rt];
    lz[rs] += lz[rt];
    lz[rt] = 0;
  }
}
void update(int rt, int l, int r, int first, int second, int v) {
  if (l >= first && r <= second) {
    lz[rt] += v;
    sum[rt] += v;
    return;
  }
  pushdown(rt);
  int mid = l + r >> 1;
  if (first <= mid) update(rt << 1, l, mid, first, second, v);
  if (second > mid) update(rt << 1 | 1, mid + 1, r, first, second, v);
  sum[rt] = min(sum[rt << 1], sum[rt << 1 | 1]);
}
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    cin >> seg[i].l >> seg[i].r >> seg[i].w, seg[i].r--;
  sort(seg + 1, seg + n + 1);
  int res = INF;
  int j = 0;
  for (int i = 1; i <= n; i++) {
    while (j < n && sum[1] <= 0) {
      j++;
      update(1, 1, m - 1, seg[j].l, seg[j].r, 1);
    }
    if (j >= i && sum[1]) res = min(res, seg[j].w - seg[i].w);
    update(1, 1, m - 1, seg[i].l, seg[i].r, -1);
  }
  cout << res << "\n";
}
int main() {
  cin.tie(0), ios::sync_with_stdio(false);
  { solve(); }
  return 0;
}

### Prompt

Generate 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;
const int N = 3e5 + 10, M = 1e6 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int P = 131;
int n, m;
int sum[M * 4], lz[M * 4];
struct node {
  int l, r, w;
  bool operator<(const node &t) const { return w < t.w; }
} seg[N];
void pushdown(int rt) {
  int ls = rt << 1, rs = rt << 1 | 1;
  if (lz[rt]) {
    sum[ls] += lz[rt];
    lz[ls] += lz[rt];
    sum[rs] += lz[rt];
    lz[rs] += lz[rt];
    lz[rt] = 0;
  }
}
void update(int rt, int l, int r, int first, int second, int v) {
  if (l >= first && r <= second) {
    lz[rt] += v;
    sum[rt] += v;
    return;
  }
  pushdown(rt);
  int mid = l + r >> 1;
  if (first <= mid) update(rt << 1, l, mid, first, second, v);
  if (second > mid) update(rt << 1 | 1, mid + 1, r, first, second, v);
  sum[rt] = min(sum[rt << 1], sum[rt << 1 | 1]);
}
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    cin >> seg[i].l >> seg[i].r >> seg[i].w, seg[i].r--;
  sort(seg + 1, seg + n + 1);
  int res = INF;
  int j = 0;
  for (int i = 1; i <= n; i++) {
    while (j < n && sum[1] <= 0) {
      j++;
      update(1, 1, m - 1, seg[j].l, seg[j].r, 1);
    }
    if (j >= i && sum[1]) res = min(res, seg[j].w - seg[i].w);
    update(1, 1, m - 1, seg[i].l, seg[i].r, -1);
  }
  cout << res << "\n";
}
int main() {
  cin.tie(0), ios::sync_with_stdio(false);
  { solve(); }
  return 0;
}

```