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 maxn = 1e5 + 100;
long long A[maxn], B[maxn], a[maxn], n, w = 1, d, b[maxn], ans;
void add(long long i, long long j, long long sw) {
  if (sw == 0) {
    for (long long y = i; y <= n; y += (y & (-y))) A[y] += j;
  } else {
    for (long long y = i; y <= n; y += (y & (-y))) B[y] += j;
  }
}
long long get(long long i, long long j) {
  long long p = 0;
  if (j == 0) {
    for (long long y = i; y; y -= (y & (-y))) p += A[y];
  } else {
    for (long long y = i; y; y -= (y & (-y))) p += B[y];
  }
  return p;
}
map<long long, long long> k;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> d;
  for (long long y = 0; y < n; y++) cin >> a[y], b[y] = a[y];
  sort(b, b + n);
  for (long long y = 0; y < n; y++) {
    if (!k[b[y]]) k[b[y]] = w, w++;
  }
  for (long long y = 0; y < n; y++) a[y] = k[a[y]];
  long long ja = 0, j = n;
  for (long long y = 0; y < n; y++)
    ja += get(n, 0) - get(a[y], 0), add(a[y], 1, 0);
  long long i = n - 2;
  while (i > -1) {
    ja -= (get(n, 0) - get(a[i + 1], 0) + get(a[i + 1] - 1, 1));
    add(a[i + 1], -1, 0);
    if (ja > d) {
      i--;
      continue;
    }
    while (j != i + 1) {
      if (ja + get(n, 0) - get(a[j - 1], 0) + get(a[j - 1] - 1, 1) <= d) {
        ja += get(n, 0) - get(a[j - 1], 0) + get(a[j - 1] - 1, 1);
        add(a[j - 1], 1, 1);
      } else
        break;
      j--;
    }
    ans += n - j;
    i--;
  }
  cout << ans;
}
