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 = 1e5 + 100;
struct bit {
  int n, a[N];
  void init(int n) {
    n += 10;
    this->n = n;
    fill(a, a + n, 0);
  }
  void add(int x, int delta) {
    x += 2;
    while (x < n) {
      a[x] += delta;
      x += x & (-x);
    }
  }
  int get(int x) {
    x += 2;
    int res = 0;
    while (x > 0) {
      res += a[x];
      x -= x & (-x);
    }
    return res;
  }
  int less(int x) { return get(x - 1); }
  int more(int x) { return get(n - 3) - get(x); }
};
int n, a[N], b[N];
long long int inv = 0;
int l, r;
bit L, R;
void init() {
  L.init(n);
  R.init(n);
  l = 0;
  r = n - 1;
  while (r >= 0) {
    inv += R.less(a[r]);
    R.add(a[r], +1);
    r--;
  }
}
void push() {
  inv += L.more(a[l]) + R.less(a[l]);
  L.add(a[l], +1);
  l++;
}
void pop() {
  r++;
  R.add(a[r], -1);
  inv -= L.more(a[r]) + R.less(a[r]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int k;
  cin >> n >> k;
  for (int i = 0; i < (int)(n); i++) cin >> a[i], b[i] = a[i];
  sort(b, b + n);
  int m = unique(b, b + n) - b;
  for (int i = 0; i < (int)(n); i++) a[i] = lower_bound(b, b + m, a[i]) - b;
  init();
  long long int ans = 0;
  for (int i = 0; i < (int)(n); i++) {
    while (r < i) pop();
    push();
    while (r < n - 1 && inv > k) pop();
    ans += n - 1 - r;
  }
  cout << ans << endl;
  return 0;
}
