#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 5;
long long n, m;
long long a[N];
long long mi[N << 2];
long long tag[N << 2];
inline long long lc(long long p) { return p << 1; }
inline long long rc(long long p) { return p << 1 | 1; }
void build(long long p, long long l, long long r) {
  if (l == r) {
    mi[p] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(lc(p), l, mid);
  build(rc(p), mid + 1, r);
  mi[p] = min(mi[lc(p)], mi[rc(p)]);
}
void add(long long p, long long l, long long r, long long ql, long long qr,
         long long d) {
  if (ql <= l && r <= qr) {
    mi[p] += d;
    tag[p] += d;
    return;
  }
  long long mid = (l + r) >> 1;
  if (tag[p]) {
    tag[lc(p)] += tag[p];
    tag[rc(p)] += tag[p];
    mi[lc(p)] += tag[p];
    mi[rc(p)] += tag[p];
    tag[p] = 0;
  }
  if (ql <= mid) add(lc(p), l, mid, ql, qr, d);
  if (qr > mid) add(rc(p), mid + 1, r, ql, qr, d);
  mi[p] = min(mi[lc(p)], mi[rc(p)]);
}
long long find_mi(long long p, long long l, long long r, long long ql,
                  long long qr) {
  if (ql <= l && r <= qr) return mi[p];
  long long mid = (l + r) >> 1;
  if (tag[p]) {
    tag[lc(p)] += tag[p];
    tag[rc(p)] += tag[p];
    mi[lc(p)] += tag[p];
    mi[rc(p)] += tag[p];
    tag[p] = 0;
  }
  long long res = 1e9;
  if (ql <= mid) res = min(res, find_mi(lc(p), l, mid, ql, qr));
  if (qr > mid) res = min(res, find_mi(rc(p), mid + 1, r, ql, qr));
  return res;
}
struct Seg {
  long long l, r, w;
} seg[N];
bool cmp(Seg s1, Seg s2) { return s1.w < s2.w; }
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    long long L, R, W;
    cin >> L >> R >> W;
    seg[i] = (Seg){L, R, W};
  }
  sort(seg + 1, seg + n + 1, cmp);
  build(1, 1, m);
  long long i = 1, j = 0, ans = 1e9;
  long long ok = 0;
  while (1) {
    if (j > n || i > n) break;
    if (!ok) {
      j++;
      if (j > n) break;
      add(1, 1, m, seg[j].l, seg[j].r - 1, 1);
      if (find_mi(1, 1, m, 1, m - 1) > 0) {
        ans = min(ans, seg[j].w - seg[i].w);
        ok = 1;
      } else {
      }
    } else {
      add(1, 1, m, seg[i].l, seg[i].r - 1, -1);
      i++;
      if (i > n) break;
      if (find_mi(1, 1, m, 1, m - 1) > 0) {
        ans = min(ans, seg[j].w - seg[i].w);
      } else {
        ok = 0;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 long long N = 1e6 + 5;
long long n, m;
long long a[N];
long long mi[N << 2];
long long tag[N << 2];
inline long long lc(long long p) { return p << 1; }
inline long long rc(long long p) { return p << 1 | 1; }
void build(long long p, long long l, long long r) {
  if (l == r) {
    mi[p] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(lc(p), l, mid);
  build(rc(p), mid + 1, r);
  mi[p] = min(mi[lc(p)], mi[rc(p)]);
}
void add(long long p, long long l, long long r, long long ql, long long qr,
         long long d) {
  if (ql <= l && r <= qr) {
    mi[p] += d;
    tag[p] += d;
    return;
  }
  long long mid = (l + r) >> 1;
  if (tag[p]) {
    tag[lc(p)] += tag[p];
    tag[rc(p)] += tag[p];
    mi[lc(p)] += tag[p];
    mi[rc(p)] += tag[p];
    tag[p] = 0;
  }
  if (ql <= mid) add(lc(p), l, mid, ql, qr, d);
  if (qr > mid) add(rc(p), mid + 1, r, ql, qr, d);
  mi[p] = min(mi[lc(p)], mi[rc(p)]);
}
long long find_mi(long long p, long long l, long long r, long long ql,
                  long long qr) {
  if (ql <= l && r <= qr) return mi[p];
  long long mid = (l + r) >> 1;
  if (tag[p]) {
    tag[lc(p)] += tag[p];
    tag[rc(p)] += tag[p];
    mi[lc(p)] += tag[p];
    mi[rc(p)] += tag[p];
    tag[p] = 0;
  }
  long long res = 1e9;
  if (ql <= mid) res = min(res, find_mi(lc(p), l, mid, ql, qr));
  if (qr > mid) res = min(res, find_mi(rc(p), mid + 1, r, ql, qr));
  return res;
}
struct Seg {
  long long l, r, w;
} seg[N];
bool cmp(Seg s1, Seg s2) { return s1.w < s2.w; }
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    long long L, R, W;
    cin >> L >> R >> W;
    seg[i] = (Seg){L, R, W};
  }
  sort(seg + 1, seg + n + 1, cmp);
  build(1, 1, m);
  long long i = 1, j = 0, ans = 1e9;
  long long ok = 0;
  while (1) {
    if (j > n || i > n) break;
    if (!ok) {
      j++;
      if (j > n) break;
      add(1, 1, m, seg[j].l, seg[j].r - 1, 1);
      if (find_mi(1, 1, m, 1, m - 1) > 0) {
        ans = min(ans, seg[j].w - seg[i].w);
        ok = 1;
      } else {
      }
    } else {
      add(1, 1, m, seg[i].l, seg[i].r - 1, -1);
      i++;
      if (i > n) break;
      if (find_mi(1, 1, m, 1, m - 1) > 0) {
        ans = min(ans, seg[j].w - seg[i].w);
      } else {
        ok = 0;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```