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 long long maxn = 1e5 + 500;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 500;
long long fen[maxn][2];
long long a[maxn];
set<long long> st;
map<long long, long long> ml;
void update(long long x, long long val, bool q) {
  x++;
  while (x < maxn) {
    fen[x][q] += val;
    x += (x & (-x));
  }
}
long long finds(long long x, bool q) {
  long long ans = 0;
  x++;
  while (x > 0) {
    ans += fen[x][q];
    x -= (x & (-x));
  }
  return ans;
}
int main() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  {
    for (long long i = 0; i < n; i++) {
      st.insert(a[i]);
    }
    long long koo = 0;
    for (set<long long>::iterator it = st.begin(); it != st.end(); it++) {
      ml[(*it)] = koo;
      koo++;
    }
    for (long long i = 0; i < n; i++) {
      a[i] = ml[a[i]];
    }
    st.clear();
    ml.clear();
  }
  long long p = 0;
  long long inv = 0;
  long long ans = 0;
  for (long long i = n - 1; i >= 0; i--) {
    update(a[i], 1, 1);
    inv += finds(a[i] - 1, 1);
  }
  if (inv <= k) {
    cout << n * (n - 1) / 2;
    return 0;
  }
  update(a[0], -1, 1);
  inv -= finds(a[0] - 1, 1);
  for (long long i = 1; i < n; i++) {
    while (inv <= k) {
      inv += finds(a[p] - 1, 1);
      inv += p - finds(a[p], 0);
      update(a[p], 1, 0);
      p++;
    }
    if (p > 0) {
      p--;
      update(a[p], -1, 0);
      inv -= finds(a[p] - 1, 1);
      inv -= p - finds(a[p], 0);
    }
    ans += p;
    update(a[i], -1, 1);
    inv -= finds(a[i] - 1, 1);
    inv -= p - finds(a[i], 0);
  }
  cout << endl << ans;
}
