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 = 100010;
long long c1[N], c2[N], a[N], l[N], r[N], dx[N], n, m, now;
int lowbit(int x) { return (x & (-x)); }
void add(long long c[], long long k, long long d) {
  for (int i = k; i <= N; i += lowbit(i)) c[i] += d;
}
long long get(long long c[], long long k) {
  long long ret = 0;
  for (int i = k; i; i -= lowbit(i)) ret += c[i];
  return ret;
}
void insert(int k) {
  now += get(c2, a[k] - 1);
  add(c1, a[k], 1);
}
void remove(int &k, int i) {
  now -= (i - get(c1, a[k]));
  add(c2, a[k], -1);
  ++k;
}
int main() {
  while (cin >> n >> m) {
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      dx[i] = a[i];
    }
    sort(dx + 1, dx + n + 1);
    int len = unique(dx + 1, dx + n + 1) - dx - 1;
    for (int i = 1; i <= n; ++i)
      a[i] = lower_bound(dx + 1, dx + n + 1, a[i]) - dx;
    l[0] = 0;
    memset(c1, 0, sizeof(c1));
    for (int i = 1; i <= n; ++i) {
      l[i] = l[i - 1] + i - 1 - get(c1, a[i]);
      add(c1, a[i], 1);
    }
    memset(c2, 0, sizeof(c2));
    for (int i = n; i >= 1; --i) {
      r[i] = r[i + 1] + get(c2, a[i] - 1);
      add(c2, a[i], 1);
    }
    memset(c1, 0, sizeof(c1));
    long long ans = 0;
    now = 0;
    int p = 1;
    for (int i = 1; i <= n; ++i) {
      insert(i);
      if (p == i) remove(p, i);
      while (p <= n && l[i] + r[p] + now > m) remove(p, i);
      ans += n - p + 1;
    }
    cout << ans << "\n";
  }
  return 0;
}
