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 INP, AM, REACHEOF;
char BUF[(1 << 12) + 1], *inp = BUF;
const long double PI = acos((long double)-1.0);
const int MN = 100111;
int n, a[MN], c[MN], f1[MN], f2[MN], sl;
long long k;
void update(int f[], int u, int k) {
  f[u] += k;
  int v = u + ((u) & (-(u)));
  if (v <= sl) update(f, v, k);
}
int get(int f[], int u) {
  if (u <= 0) return 0;
  return f[u] + get(f, u - ((u) & (-(u))));
}
int main() {
  while (cin >> n >> k) {
    for (int i = (1), _b = (n); i <= _b; i++) {
      scanf("%d", &a[i]);
      c[i] = a[i];
    }
    sort(c + 1, c + n + 1);
    sl = unique(c + 1, c + n + 1) - c - 1;
    for (int i = (1), _b = (n); i <= _b; i++)
      a[i] = lower_bound(c + 1, c + sl + 1, a[i]) - c;
    long long inv = 0;
    memset(f1, 0, sizeof f1);
    memset(f2, 0, sizeof f2);
    for (int i = (n), _b = (1); i >= _b; i--) {
      inv += get(f2, a[i] - 1);
      update(f2, a[i], 1);
    }
    if (inv <= k) {
      cout << n * (n - 1LL) / 2 << endl;
      continue;
    }
    int r = 2;
    long long res = 0;
    update(f2, a[1], -1);
    for (int l = (1), _b = (n); l <= _b; l++) {
      if (l > 1) {
        inv += l - 1 - get(f1, a[l]);
        inv += get(f2, a[l] - 1);
      }
      update(f1, a[l], 1);
      while (r < n && inv > k) {
        inv -= l - get(f1, a[r]);
        inv -= get(f2, a[r] - 1);
        update(f2, a[r], -1);
        ++r;
      }
      if (inv <= k) {
        res += n - r + 1;
      }
    }
    cout << res << endl;
  }
  return 0;
}
