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>
const long double eps = 1e-9;
const double pi = acos(-1.0);
const long long inf = 1e18;
using namespace std;
struct summator {
  int a[111111];
  summator() { clear(); }
  void clear() {
    for (int i = 0; i < 111111; i++) a[i] = 0;
  }
  void modify(int pos, int val) {
    while (pos < 111111) {
      a[pos] += val;
      pos = 2 * pos - (pos & (pos - 1));
    }
  }
  int fsum(int pos) {
    int res = 0;
    while (pos != 0) {
      res += a[pos];
      pos = pos & (pos - 1);
    }
    return res;
  }
};
int n, a[111111], b[111111], ll, rr;
long long k, num = 0, ans = 0;
map<int, int> mm;
summator s1, s2;
void compress() {
  sort(b + 1, b + n + 1);
  int id = 1;
  mm[b[1]] = id;
  for (int i = 2; i <= n; i++) {
    if (b[i - 1] == b[i]) continue;
    ++id;
    mm[b[i]] = id;
  }
  for (int i = 1; i <= n; i++) a[i] = mm[a[i]];
}
void inc_l() {
  ++ll;
  num += s1.fsum(100111) - s1.fsum(a[ll]);
  num += s2.fsum(a[ll] - 1);
  s1.modify(a[ll], 1);
}
void dec_l() {
  num -= s1.fsum(100111) - s1.fsum(a[ll]);
  num -= s2.fsum(a[ll] - 1);
  s1.modify(a[ll], -1);
  --ll;
}
void dec_r() {
  --rr;
  num += s2.fsum(a[rr] - 1);
  num += s1.fsum(100111) - s1.fsum(a[rr]);
  s2.modify(a[rr], 1);
}
int main(int argc, const char* argv[]) {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
  compress();
  ll = 0;
  rr = n + 1;
  dec_r();
  while (ll + 1 < rr) {
    inc_l();
    if (num > k) {
      dec_l();
      break;
    }
  }
  if (ll == 0) {
    cout << 0;
    return 0;
  }
  while (rr != 1) {
    dec_r();
    ans += ll;
    if (ll == rr) dec_l();
    while (num > k) {
      dec_l();
      if (ll == 0) {
        cout << ans;
        return 0;
      }
    }
  }
  cout << ans;
  return 0;
}
