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 long MOD = 1e9 + 7;
using namespace std;
long long n, k;
vector<long long> arr;
int fen[2][100005];
void remove(long long x, int type) {
  while (x <= (n + 1)) {
    fen[type][x]--;
    x += (x & -x);
  }
  return;
}
void update(long long x, int type) {
  while (x <= (n + 1)) {
    fen[type][x]++;
    x += (x & -x);
  }
  return;
}
long long query(long long x, int type) {
  long long res = 0;
  while (x > 0) {
    res += fen[type][x];
    x -= (x & -x);
  }
  return res;
}
void data_compress() {
  vector<long long> temp = arr;
  sort(temp.begin(), temp.end());
  for (int i = 0; i < n; i++)
    arr[i] = (lower_bound(temp.begin(), temp.end(), arr[i]) - temp.begin()) + 1;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  cin.tie(NULL);
  cout.tie(NULL);
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 100005; j++) fen[i][j] = 0;
  }
  cin >> n >> k;
  arr.resize(n, 0);
  for (int i = 0; i < n; i++) cin >> arr[i];
  data_compress();
  vector<long long> right(n, 0);
  long long r = n - 1, cur = 0;
  while (r >= 0) {
    long long inv = query(arr[r], 1);
    right[r] = inv;
    cur += inv;
    if (cur > k) {
      cur -= inv;
      r++;
      break;
    }
    update(arr[r] + 1, 1);
    r--;
  }
  if (r == -1) {
    cout << n * (n - 1) / 2;
    return 0;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    cur += query(arr[i], 1);
    cur += (i - query(arr[i] + 1, 0));
    update(arr[i] + 1, 0);
    while ((cur > k) && (r < n)) {
      long long res = right[r];
      res += (i + 1 - query(arr[r] + 1, 0));
      cur -= res;
      remove(arr[r] + 1, 1);
      r++;
    }
    if (r == n) break;
    if (cur <= k) ans += (n - r);
  }
  cout << ans;
  return 0;
}
