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;
vector<int> L, P;
int l, size;
void Stworz(vector<int>& T, int n) {
  int p = 1;
  while (p < n) p *= 2;
  l = p;
  size = 2 * p - 1;
  for (int i = 0; i <= size; ++i) T.push_back(0);
}
void Dodaj(vector<int>& T, int a, int x) {
  a += l;
  T[a] += x;
  a /= 2;
  while (a != 0) {
    T[a] = T[2 * a] + T[2 * a + 1];
    a /= 2;
  }
}
int Ile(vector<int>& T, int a, int b) {
  if (a > b) return 0;
  a += l;
  b += l;
  int s = T[a];
  if (a != b) s += T[b];
  while (a / 2 != b / 2) {
    if (a % 2 == 0) s += T[a + 1];
    if (b % 2 == 1) s += T[b - 1];
    a /= 2;
    b /= 2;
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<int> b = a;
  sort(b.begin(), b.end());
  map<int, int> M;
  int nr = 0;
  for (int i = 0; i < n; ++i) {
    if (i == 0 || b[i] != b[i - 1]) {
      M[b[i]] = nr;
      ++nr;
    }
  }
  Stworz(L, nr);
  Stworz(P, nr);
  long long inv = 0, odp = 0;
  int l = -1, p = n - 1;
  for (p; p >= 1; --p) {
    Dodaj(P, M[a[p]], 1);
    inv += Ile(P, 0, M[a[p]] - 1);
  }
  for (p = 1; p < n; ++p) {
    while (l + 1 < p && inv <= k) {
      ++l;
      Dodaj(L, M[a[l]], 1);
      inv += Ile(L, M[a[l]] + 1, nr - 1) + Ile(P, 0, M[a[l]] - 1);
    }
    if (inv > k && l != -1) {
      Dodaj(L, M[a[l]], -1);
      inv -= Ile(L, M[a[l]] + 1, nr - 1) + Ile(P, 0, M[a[l]] - 1);
      --l;
    }
    if (inv <= k && l != -1) odp += l + 1;
    Dodaj(P, M[a[p]], -1);
    inv -= Ile(L, M[a[p]] + 1, nr - 1) + Ile(P, 0, M[a[p]] - 1);
  }
  cout << odp;
  return 0;
}
