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>
using namespace std;
int n, m;
long long k;
int a[100005], s[100005], r[100005];
struct BIT {
  int c[100005];
  void add(int x, int y) {
    for (int i = x; i <= m; i += (i & -i)) {
      c[i] += y;
    }
  }
  long long get(int x) {
    long long ret = 0;
    for (int i = x; i > 0; i -= (i & -i)) {
      ret += c[i];
    }
    return ret;
  }
} f0, f1;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    s[i] = a[i];
  }
  sort(s, s + n);
  m = unique(s, s + n) - s;
  for (int i = 0; i < n; ++i) {
    a[i] = lower_bound(s, s + m, a[i]) - s + 1;
  }
  for (int i = n - 1; i >= 0; --i) {
    r[i] = f0.get(a[i] - 1) + (i < n - 1 ? r[i + 1] : 0);
    f0.add(a[i], 1);
  }
  for (int i = 0; i < n; ++i) {
    f0.add(a[i], -1);
  }
  f0.add(a[0], 1);
  int pos = n;
  long long now = 0;
  for (int i = n - 1; i >= 1; --i) {
    f1.add(a[i], 1);
    if (r[i] + f1.get(a[0] - 1) > k) {
      f1.add(a[i], -1);
      break;
    }
    pos = i;
    now = r[i] + f1.get(a[0] - 1);
  }
  if (pos == 1) {
    cout << 1ll * n * (n - 1) / 2;
    return 0;
  }
  if (pos == n) {
    puts("0");
    return 0;
  }
  long long ans = n - pos;
  for (int i = 1; i < n - 1; ++i) {
    now += f0.get(m) - f0.get(a[i]) + f1.get(a[i] - 1);
    f0.add(a[i], 1);
    while (now > k && pos < n) {
      now -= f0.get(m) - f0.get(a[pos]) + f1.get(a[pos] - 1);
      f1.add(a[pos], -1);
      pos++;
    }
    if (pos == n && now > k) {
      break;
    }
    ans += n - pos;
  }
  cout << ans << endl;
  return 0;
}
