The Little Elephant has array a, consisting of n positive integers, indexed from 1 to n. Let's denote the number with index i as ai.

The Little Elephant wants to count, how many pairs of integers l and r are there, such that 1 ≤ l < r ≤ n and sequence b = a1a2... alarar + 1... an has no more than k inversions. 

An inversion in sequence b is a pair of elements of the sequence b, that change their relative order after a stable sorting of the sequence. In other words, an inversion is a pair of integers i and j, such that 1 ≤ i < j ≤ |b| and bi > bj, where |b| is the length of sequence b, and bj is its j-th element.

Help the Little Elephant and count the number of the described pairs.

Input

The first line contains two integers n and k (2 ≤ n ≤ 105, 0 ≤ k ≤ 1018) — the size of array a and the maximum allowed number of inversions respectively. The next line contains n positive integers, separated by single spaces, a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of array a.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output

In a single line print a single number — the answer to the problem.

Examples

Input

3 1
1 3 2


Output

3


Input

5 2
1 3 2 1 7


Output

6
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
int a, b, c, d, i, j, n, m, k;
long long lim;
int mas[100002], num[100002];
int fen[100002], fen2[100002];
long long pr[100002], suf[100002];
inline void upd(int first, int val) {
  for (; first < k; first |= first + 1) fen[first] += val;
}
inline int sum(int first) {
  int ans = 0;
  for (; first >= 0; first = (first & (first + 1)) - 1) ans += fen[first];
  return ans;
}
inline int sum(int l, int r) {
  if (l > r) return 0;
  return sum(r) - sum(l - 1);
}
inline void upd2(int first, int val) {
  for (; first < k; first |= first + 1) fen2[first] += val;
}
inline int sum2(int first) {
  int ans = 0;
  for (; first >= 0; first = (first & (first + 1)) - 1) ans += fen2[first];
  return ans;
}
inline int sum2(int l, int r) {
  if (l > r) return 0;
  return sum2(r) - sum2(l - 1);
}
int main() {
  scanf("%d%I64d", &n, &lim);
  for (int _n((n)-1), i(0); i <= _n; i++)
    scanf("%d", &mas[i]), num[k++] = mas[i];
  sort(num, num + k);
  k = unique(num, num + k) - num;
  for (int _n((n)-1), i(0); i <= _n; i++)
    mas[i] = lower_bound(num, num + k, mas[i]) - num;
  long long cur = 0;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    cur += sum(mas[i] + 1, k - 1);
    pr[i] = cur;
    upd(mas[i], 1);
  }
  memset((fen), 0, sizeof(fen));
  cur = 0;
  for (int i = (n - 1), _b = (0); i >= _b; i--) {
    cur += sum(mas[i] - 1);
    suf[i] = cur;
    upd(mas[i], 1);
  }
  memset((fen), 0, sizeof(fen));
  a = n - 1;
  cur = 0;
  long long ans = 0;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    upd(mas[i], 1);
    cur += sum2(mas[i] - 1);
    while (a > i) {
      long long diff = sum(mas[a] + 1, k - 1);
      long long now = pr[i] + suf[a] + diff + cur;
      if (now > lim) break;
      cur += diff;
      upd2(mas[a], 1);
      --a;
    }
    while (a < i) {
      long long diff = sum(mas[a + 1] + 1, k - 1);
      cur -= diff;
      upd2(mas[a + 1], -1);
      ++a;
    }
    while (a < n - 1) {
      long long now = pr[i] + suf[a + 1] + cur;
      if (now <= lim) break;
      long long diff = sum(mas[a + 1] + 1, k - 1);
      cur -= diff;
      upd2(mas[a + 1], -1);
      ++a;
    }
    ans += n - 1 - a;
  }
  cout << ans << endl;
}
