#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct lazyseg {
  int base = INT_MAX;
  int comb(int a, int b) { return min(a, b); }
  struct node {
    int val, lazy;
  };
  vector<node> st;
  lazyseg(int n) : st(4 * n + 10) {}
  void puttag(int k, int l, int r, int x) {
    st[k].val += x;
    st[k].lazy += x;
  }
  void pushdown(int k, int l, int r) {
    if (l != r) {
      int m = (l + r) / 2;
      puttag(2 * k, l, m, st[k].lazy);
      puttag(2 * k + 1, m + 1, r, st[k].lazy);
    }
    st[k].lazy = 0;
  }
  void update(int a, int b, int k, int l, int r, int x) {
    if (a <= l && r <= b) {
      puttag(k, l, r, x);
      return;
    }
    pushdown(k, l, r);
    int m = (l + r) / 2;
    if (a <= m) update(a, b, 2 * k, l, m, x);
    if (b > m) update(a, b, 2 * k + 1, m + 1, r, x);
    st[k].val = comb(st[2 * k].val, st[2 * k + 1].val);
  }
  int get(int a, int b, int k, int l, int r) {
    if (a <= l && r <= b) return st[k].val;
    pushdown(k, l, r);
    int m = (l + r) / 2;
    int ret = base;
    if (a <= m) ret = comb(ret, get(a, b, 2 * k, l, m));
    if (b > m) ret = comb(ret, get(a, b, 2 * k + 1, m + 1, r));
    return ret;
  }
  void build(vector<int> &v, int k, int l, int r) {
    if (l == r) {
      st[k].val = v[l];
      return;
    }
    int m = (l + r) / 2;
    build(v, 2 * k, l, m);
    build(v, 2 * k + 1, m + 1, r);
    st[k].val = comb(st[2 * k].val, st[2 * k + 1].val);
  }
};
const int mx = 2e6 + 5;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  m *= 2;
  vector<array<int, 3>> v;
  for (int q = 0; q < n; q++) {
    int a, b, c;
    cin >> a >> b >> c;
    a *= 2;
    b *= 2;
    v.push_back({c, a, b});
  }
  sort(v.begin(), v.end());
  lazyseg st(mx);
  int lo = -1;
  int ans = INT_MAX;
  for (int q = 0; q < n; q++) {
    st.update(v[q][1], v[q][2], 1, 0, mx - 1, 1);
    if (lo < 0) lo++;
    while (lo < n && st.get(2, m, 1, 0, mx - 1) > 0) {
      if (lo >= 0) {
        st.update(v[lo][1], v[lo][2], 1, 0, mx - 1, -1);
      }
      if (st.get(2, m, 1, 0, mx - 1) > 0) {
        lo++;
        continue;
      }
      if (lo >= 0) {
        st.update(v[lo][1], v[lo][2], 1, 0, mx - 1, 1);
      }
      break;
    }
    if (st.get(2, m, 1, 0, mx - 1) > 0) ans = min(ans, v[q][0] - v[lo][0]);
  }
  cout << ans << '\n';
  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>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct lazyseg {
  int base = INT_MAX;
  int comb(int a, int b) { return min(a, b); }
  struct node {
    int val, lazy;
  };
  vector<node> st;
  lazyseg(int n) : st(4 * n + 10) {}
  void puttag(int k, int l, int r, int x) {
    st[k].val += x;
    st[k].lazy += x;
  }
  void pushdown(int k, int l, int r) {
    if (l != r) {
      int m = (l + r) / 2;
      puttag(2 * k, l, m, st[k].lazy);
      puttag(2 * k + 1, m + 1, r, st[k].lazy);
    }
    st[k].lazy = 0;
  }
  void update(int a, int b, int k, int l, int r, int x) {
    if (a <= l && r <= b) {
      puttag(k, l, r, x);
      return;
    }
    pushdown(k, l, r);
    int m = (l + r) / 2;
    if (a <= m) update(a, b, 2 * k, l, m, x);
    if (b > m) update(a, b, 2 * k + 1, m + 1, r, x);
    st[k].val = comb(st[2 * k].val, st[2 * k + 1].val);
  }
  int get(int a, int b, int k, int l, int r) {
    if (a <= l && r <= b) return st[k].val;
    pushdown(k, l, r);
    int m = (l + r) / 2;
    int ret = base;
    if (a <= m) ret = comb(ret, get(a, b, 2 * k, l, m));
    if (b > m) ret = comb(ret, get(a, b, 2 * k + 1, m + 1, r));
    return ret;
  }
  void build(vector<int> &v, int k, int l, int r) {
    if (l == r) {
      st[k].val = v[l];
      return;
    }
    int m = (l + r) / 2;
    build(v, 2 * k, l, m);
    build(v, 2 * k + 1, m + 1, r);
    st[k].val = comb(st[2 * k].val, st[2 * k + 1].val);
  }
};
const int mx = 2e6 + 5;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  m *= 2;
  vector<array<int, 3>> v;
  for (int q = 0; q < n; q++) {
    int a, b, c;
    cin >> a >> b >> c;
    a *= 2;
    b *= 2;
    v.push_back({c, a, b});
  }
  sort(v.begin(), v.end());
  lazyseg st(mx);
  int lo = -1;
  int ans = INT_MAX;
  for (int q = 0; q < n; q++) {
    st.update(v[q][1], v[q][2], 1, 0, mx - 1, 1);
    if (lo < 0) lo++;
    while (lo < n && st.get(2, m, 1, 0, mx - 1) > 0) {
      if (lo >= 0) {
        st.update(v[lo][1], v[lo][2], 1, 0, mx - 1, -1);
      }
      if (st.get(2, m, 1, 0, mx - 1) > 0) {
        lo++;
        continue;
      }
      if (lo >= 0) {
        st.update(v[lo][1], v[lo][2], 1, 0, mx - 1, 1);
      }
      break;
    }
    if (st.get(2, m, 1, 0, mx - 1) > 0) ans = min(ans, v[q][0] - v[lo][0]);
  }
  cout << ans << '\n';
  return 0;
}

```