#include <bits/stdc++.h>
using namespace std;
const int N = 3 * 1e5 + 100, M = 1e6 + 100;
int n, m, MIN[M * 4], tag[M * 4];
struct node {
  int l, r, w;
} sh[N];
bool operator<(node a, node b) { return a.w < b.w; }
void pushup(int x) { MIN[x] = min(MIN[x + x], MIN[x + x + 1]); }
void pushdown(int x) {
  if (tag[x]) {
    MIN[x + x] += tag[x];
    MIN[x + x + 1] += tag[x];
    tag[x + x] += tag[x];
    tag[x + x + 1] += tag[x];
    tag[x] = 0;
  }
}
void change(int x, int l, int r, int ll, int rr, int v) {
  if (ll <= l && rr >= r) {
    MIN[x] += v;
    tag[x] += v;
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(x);
  if (ll <= mid) change(x + x, l, mid, ll, rr, v);
  if (rr > mid) change(x + x + 1, mid + 1, r, ll, rr, v);
  pushup(x);
}
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
void add(int x, int v) {
  if (!x) return;
  change(1, 1, m - 1, sh[x].l, sh[x].r - 1, v);
}
signed main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) sh[i] = (node){read(), read(), read()};
  sort(sh + 1, sh + 1 + n);
  int l = n, r = n, ans = 0x3f3f3f3f;
  add(n, 1);
  if (MIN[1] != 0) ans = 0;
  while (l <= r && l >= 1) {
    if (MIN[1] == 0)
      add(--l, 1);
    else
      add(r--, -1);
    if (MIN[1] != 0) ans = min(ans, sh[r].w - sh[l].w);
  }
  printf("%d\n", ans);
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 3 * 1e5 + 100, M = 1e6 + 100;
int n, m, MIN[M * 4], tag[M * 4];
struct node {
  int l, r, w;
} sh[N];
bool operator<(node a, node b) { return a.w < b.w; }
void pushup(int x) { MIN[x] = min(MIN[x + x], MIN[x + x + 1]); }
void pushdown(int x) {
  if (tag[x]) {
    MIN[x + x] += tag[x];
    MIN[x + x + 1] += tag[x];
    tag[x + x] += tag[x];
    tag[x + x + 1] += tag[x];
    tag[x] = 0;
  }
}
void change(int x, int l, int r, int ll, int rr, int v) {
  if (ll <= l && rr >= r) {
    MIN[x] += v;
    tag[x] += v;
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(x);
  if (ll <= mid) change(x + x, l, mid, ll, rr, v);
  if (rr > mid) change(x + x + 1, mid + 1, r, ll, rr, v);
  pushup(x);
}
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
void add(int x, int v) {
  if (!x) return;
  change(1, 1, m - 1, sh[x].l, sh[x].r - 1, v);
}
signed main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) sh[i] = (node){read(), read(), read()};
  sort(sh + 1, sh + 1 + n);
  int l = n, r = n, ans = 0x3f3f3f3f;
  add(n, 1);
  if (MIN[1] != 0) ans = 0;
  while (l <= r && l >= 1) {
    if (MIN[1] == 0)
      add(--l, 1);
    else
      add(r--, -1);
    if (MIN[1] != 0) ans = min(ans, sh[r].w - sh[l].w);
  }
  printf("%d\n", ans);
}

```