#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
const long long mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template <typename T>
inline void read(T &a) {
  char c = getchar();
  T x = 0, f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  a = f * x;
}
int gcd(int a, int b) { return (b > 0) ? gcd(b, a % b) : a; }
struct P {
  int l, r, w;
} a[N];
struct Node {
  int l, r;
  int x, add;
} tr[N * 4];
int n, m;
bool cmp(P A, P B) { return A.w < B.w; }
void push(int u) { tr[u].x = min(tr[u << 1].x, tr[u << 1 | 1].x); }
void down(int u) {
  if (tr[u].add != 0) {
    int &k = tr[u].add;
    tr[u << 1].x += k;
    tr[u << 1 | 1].x += k;
    tr[u << 1].add += k;
    tr[u << 1 | 1].add += k;
    k = 0;
  }
}
void build(int u, int l, int r) {
  tr[u] = {l, r, 0, 0};
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
}
void modify(int u, int l, int r, int k) {
  if (tr[u].l >= l && tr[u].r <= r) {
    tr[u].x += k, tr[u].add += k;
    return;
  }
  down(u);
  int mid = (tr[u].l + tr[u].r) >> 1;
  if (l <= mid) modify(u << 1, l, r, k);
  if (r > mid) modify(u << 1 | 1, l, r, k);
  push(u);
}
int main() {
  read(n), read(m);
  m--;
  build(1, 1, m);
  for (int i = 1; i <= n; i++) read(a[i].l), read(a[i].r), read(a[i].w);
  sort(a + 1, a + 1 + n, cmp);
  int l = 0, r = 0, ans = INF;
  while (r <= n) {
    while (r + 1 <= n && tr[1].x == 0) r++, modify(1, a[r].l, a[r].r - 1, 1);
    if (!tr[1].x) break;
    while (l + 1 <= r && tr[1].x > 0) l++, modify(1, a[l].l, a[l].r - 1, -1);
    ans = min(ans, a[r].w - a[l].w);
  }
  printf("%d\n", ans);
  ;
  return 0;
}

### Prompt

Your task is to create 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>
#pragma GCC optimize(3)
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
const long long mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template <typename T>
inline void read(T &a) {
  char c = getchar();
  T x = 0, f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  a = f * x;
}
int gcd(int a, int b) { return (b > 0) ? gcd(b, a % b) : a; }
struct P {
  int l, r, w;
} a[N];
struct Node {
  int l, r;
  int x, add;
} tr[N * 4];
int n, m;
bool cmp(P A, P B) { return A.w < B.w; }
void push(int u) { tr[u].x = min(tr[u << 1].x, tr[u << 1 | 1].x); }
void down(int u) {
  if (tr[u].add != 0) {
    int &k = tr[u].add;
    tr[u << 1].x += k;
    tr[u << 1 | 1].x += k;
    tr[u << 1].add += k;
    tr[u << 1 | 1].add += k;
    k = 0;
  }
}
void build(int u, int l, int r) {
  tr[u] = {l, r, 0, 0};
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
}
void modify(int u, int l, int r, int k) {
  if (tr[u].l >= l && tr[u].r <= r) {
    tr[u].x += k, tr[u].add += k;
    return;
  }
  down(u);
  int mid = (tr[u].l + tr[u].r) >> 1;
  if (l <= mid) modify(u << 1, l, r, k);
  if (r > mid) modify(u << 1 | 1, l, r, k);
  push(u);
}
int main() {
  read(n), read(m);
  m--;
  build(1, 1, m);
  for (int i = 1; i <= n; i++) read(a[i].l), read(a[i].r), read(a[i].w);
  sort(a + 1, a + 1 + n, cmp);
  int l = 0, r = 0, ans = INF;
  while (r <= n) {
    while (r + 1 <= n && tr[1].x == 0) r++, modify(1, a[r].l, a[r].r - 1, 1);
    if (!tr[1].x) break;
    while (l + 1 <= r && tr[1].x > 0) l++, modify(1, a[l].l, a[l].r - 1, -1);
    ans = min(ans, a[r].w - a[l].w);
  }
  printf("%d\n", ans);
  ;
  return 0;
}

```