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;
const int N = 100005;
int n, a[N];
long long k;
vector<int> t[N + N];
int big(int l, int r, int x) {
  int ret = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      ret += (int)t[l].size() -
             (upper_bound(t[l].begin(), t[l].end(), x) - t[l].begin());
      ++l;
    }
    if (r & 1) {
      --r;
      ret += (int)t[r].size() -
             (upper_bound(t[r].begin(), t[r].end(), x) - t[r].begin());
    }
  }
  return ret;
}
int small(int l, int r, int x) {
  int ret = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      ret += lower_bound(t[l].begin(), t[l].end(), x) - t[l].begin();
      ++l;
    }
    if (r & 1) {
      --r;
      ret += lower_bound(t[r].begin(), t[r].end(), x) - t[r].begin();
    }
  }
  return ret;
}
int main() {
  scanf("%d %lld", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    t[n + i].push_back(a[i]);
  }
  for (int i = n - 1; i; --i) {
    merge(t[i << 1].begin(), t[i << 1].end(), t[i << 1 | 1].begin(),
          t[i << 1 | 1].end(), back_inserter(t[i]));
  }
  long long curr = 0;
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    curr += (long long)big(0, i, a[i]);
  }
  for (int i = 1, j = 1; i < n; ++i) {
    while (j < n - 1 and curr > k) {
      long long out = (long long)(big(0, i, a[j]) + small(j, n, a[j]));
      curr -= out, ++j;
    }
    if (curr <= k) {
      ans += (long long)(n - j);
    }
    if (i == j) {
      ++j;
    } else {
      long long add = (long long)(big(0, i, a[i]) + small(j, n, a[i]));
      curr += add;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
