#include <bits/stdc++.h>
using namespace std;
int read() {
  int a = 0;
  int f = 0;
  char p = getchar();
  while (!isdigit(p)) {
    f |= p == '-';
    p = getchar();
  }
  while (isdigit(p)) {
    a = (a << 3) + (a << 1) + (p ^ 48);
    p = getchar();
  }
  return f ? -a : a;
}
const int INF = 998244353;
int n, m, ans;
struct node {
  int l, r, w;
  bool operator<(const node& p) const { return w < p.w; }
} seg[1000050];
int mn[4000050];
int tag[4000050];
void pushdown(int u) {
  mn[u << 1] += tag[u];
  tag[u << 1] += tag[u];
  mn[u << 1 | 1] += tag[u];
  tag[u << 1 | 1] += tag[u];
  tag[u] = 0;
}
void update(int u) { mn[u] = min(mn[u << 1], mn[u << 1 | 1]); }
void modify(int u, int l, int r, int L, int R, int V) {
  if (L <= l && r <= R) {
    mn[u] += V;
    tag[u] += V;
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(u);
  if (L <= mid) modify(u << 1, l, mid, L, R, V);
  if (R > mid) modify(u << 1 | 1, mid + 1, r, L, R, V);
  update(u);
}
int main() {
  ans = INF;
  n = read();
  m = read();
  for (int i = 1; i <= n; ++i) {
    seg[i].l = read();
    seg[i].r = read();
    seg[i].w = read();
  }
  sort(seg + 1, seg + n + 1);
  for (int i = 1, now = 0; i <= n; ++i) {
    while (mn[1] == 0 && now < n) {
      now++;
      modify(1, 2, m, seg[now].l + 1, seg[now].r, 1);
    }
    if (mn[1])
      ans = min(ans, seg[now].w - seg[i].w);
    else
      break;
    modify(1, 2, m, seg[i].l + 1, seg[i].r, -1);
  }
  printf("%d", ans);
  return 0;
}

### Prompt

Please formulate 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;
int read() {
  int a = 0;
  int f = 0;
  char p = getchar();
  while (!isdigit(p)) {
    f |= p == '-';
    p = getchar();
  }
  while (isdigit(p)) {
    a = (a << 3) + (a << 1) + (p ^ 48);
    p = getchar();
  }
  return f ? -a : a;
}
const int INF = 998244353;
int n, m, ans;
struct node {
  int l, r, w;
  bool operator<(const node& p) const { return w < p.w; }
} seg[1000050];
int mn[4000050];
int tag[4000050];
void pushdown(int u) {
  mn[u << 1] += tag[u];
  tag[u << 1] += tag[u];
  mn[u << 1 | 1] += tag[u];
  tag[u << 1 | 1] += tag[u];
  tag[u] = 0;
}
void update(int u) { mn[u] = min(mn[u << 1], mn[u << 1 | 1]); }
void modify(int u, int l, int r, int L, int R, int V) {
  if (L <= l && r <= R) {
    mn[u] += V;
    tag[u] += V;
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(u);
  if (L <= mid) modify(u << 1, l, mid, L, R, V);
  if (R > mid) modify(u << 1 | 1, mid + 1, r, L, R, V);
  update(u);
}
int main() {
  ans = INF;
  n = read();
  m = read();
  for (int i = 1; i <= n; ++i) {
    seg[i].l = read();
    seg[i].r = read();
    seg[i].w = read();
  }
  sort(seg + 1, seg + n + 1);
  for (int i = 1, now = 0; i <= n; ++i) {
    while (mn[1] == 0 && now < n) {
      now++;
      modify(1, 2, m, seg[now].l + 1, seg[now].r, 1);
    }
    if (mn[1])
      ans = min(ans, seg[now].w - seg[i].w);
    else
      break;
    modify(1, 2, m, seg[i].l + 1, seg[i].r, -1);
  }
  printf("%d", ans);
  return 0;
}

```