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:256000000")
using namespace std;
const int maxN = 110000;
int n, a[maxN];
long long k;
int t1[maxN], t2[maxN];
void update(int t[], int k, int value) {
  for (int i = k; i < maxN; i |= (i + 1)) {
    t[i] += value;
  }
}
int get(int t[], int k) {
  int res = 0;
  for (int i = k; i >= 0; i = (i & (i + 1)) - 1) {
    res += t[i];
  }
  return res;
}
int get(int t[], int l, int r) {
  if (l > r) {
    return 0;
  }
  return get(t, r) - get(t, l - 1);
}
int main() {
  scanf("%d%I64d", &n, &k);
  vector<int> b;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    b.push_back(a[i]);
  }
  sort(b.begin(), b.end());
  b.resize(unique(b.begin(), b.end()) - b.begin());
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
    ++a[i];
  }
  long long res = 0;
  long long total = 0;
  update(t1, a[1], 1);
  for (int i = n; i > 1; --i) {
    total += get(t2, 1, a[i] - 1);
    total += get(t1, a[i] + 1, n);
    update(t2, a[i], 1);
  }
  int bound = 2;
  for (int i = 1; i < n; ++i) {
    while (bound <= n && total > k) {
      total -= get(t1, a[bound] + 1, n);
      total -= get(t2, 1, a[bound] - 1);
      update(t2, a[bound], -1);
      ++bound;
    }
    if (bound == n + 1) {
      break;
    }
    res += (n + 1 - bound);
    if (i + 1 == n) {
      continue;
    }
    if (bound == i + 1) {
      total -= get(t1, a[bound] + 1, n);
      total -= get(t2, 1, a[bound] - 1);
      update(t2, a[bound], -1);
      ++bound;
    }
    {
      total += get(t1, a[i + 1] + 1, n);
      total += get(t2, 1, a[i + 1] - 1);
      update(t1, a[i + 1], 1);
    }
  }
  printf("%I64d\n", res);
  return 0;
}
