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;
long long f1[100010], f2[100010];
int n, m, k;
int l, r, num;
int a, b, c;
long long sum, ni;
long long arr[100010], dd[100010];
map<long long, int> bh;
int lowbit(int x) { return x & (x ^ (x - 1)); }
void add1(int w, int v) {
  int i;
  for (i = w; i <= num; i += lowbit(i)) f1[i] += v;
}
void add2(int w, int v) {
  int i;
  for (i = w; i <= num; i += lowbit(i)) f2[i] += v;
}
long long que1(int w) {
  long long res = 0;
  for (int i = w; i >= 1; i -= lowbit(i)) res += f1[i];
  return res;
}
long long que2(int w) {
  long long res = 0;
  for (int i = w; i >= 1; i -= lowbit(i)) res += f2[i];
  return res;
}
int main() {
  cin >> n >> k;
  for (a = 1; a <= n; a++) {
    cin >> arr[a];
    dd[a] = arr[a];
  }
  sort(dd + 1, dd + n + 1);
  for (a = 1; a <= n; a++)
    if (bh[dd[a]] == 0) {
      num++;
      bh[dd[a]] = num;
    }
  l = 1;
  for (r = n; r >= 2; r--) {
    ni += que2(bh[arr[r]] - 1);
    add2(bh[arr[r]], 1);
  }
  r = 2;
  for (l = 1; l < n; l++) {
    ni += que1(num) - que1(bh[arr[l]]);
    ni += que2(bh[arr[l]] - 1);
    add1(bh[arr[l]], 1);
    if (r == l) {
      add2(bh[arr[r]], -1);
      ni -= que1(num) - que1(bh[arr[r]]);
      ni -= que2(bh[arr[r]] - 1);
      r++;
    }
    while (ni > k && r <= n) {
      add2(bh[arr[r]], -1);
      ni -= que1(num) - que1(bh[arr[r]]);
      ni -= que2(bh[arr[r]] - 1);
      r++;
    }
    if (r > n) break;
    sum += (n - r + 1);
  }
  cout << sum << endl;
}
