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 = (int)1e5 + 123;
const long long INF = (long long)1e18 + 123;
const int inf = (int)1e9 + 123;
const int MOD = (int)1e9 + 7;
void megaRandom() {
  unsigned int FOR;
  asm("rdtsc" : "=A"(FOR));
  srand(FOR);
}
int n, a[N];
long long k;
int t[3][(int)5e6], L[3][(int)5e6], R[3][(int)5e6], Sz[3];
void upd(int type, int pos, int val, int v = 1, int tl = 1, int tr = 1e9) {
  if (tl == tr) {
    t[type][v] += val;
    return;
  }
  int mid = (tl + tr) >> 1;
  if (pos <= mid) {
    if (!L[type][v]) L[type][v] = ++Sz[type];
    upd(type, pos, val, L[type][v], tl, mid);
  } else {
    if (!R[type][v]) R[type][v] = ++Sz[type];
    upd(type, pos, val, R[type][v], mid + 1, tr);
  }
  t[type][v] = t[type][L[type][v]] + t[type][R[type][v]];
}
int get(int type, int l, int r, int v = 1, int tl = 1, int tr = 1e9) {
  if (!v || l > r || tl > tr || tl > r || l > tr) return 0;
  if (l <= tl && tr <= r) return t[type][v];
  int mid = (tl + tr) >> 1;
  return get(type, l, r, L[type][v], tl, mid) +
         get(type, l, r, R[type][v], mid + 1, tr);
}
int main() {
  megaRandom();
  Sz[1] = Sz[2] = 1;
  scanf("%d%I64d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int r = 2;
  long long every_suff = 0;
  for (int i = n; i >= 2; i--) {
    upd(2, a[i], 1);
    every_suff += get(2, 1, a[i] - 1);
  }
  long long res = 0, every = 0;
  for (int l = 1; l < n; l++) {
    if (r == l) {
      upd(2, a[r], -1);
      every_suff -= get(2, 1, a[r] - 1);
      every -= get(1, a[r] + 1, 1e9);
      r++;
    }
    every += get(1, a[l] + 1, 1e9);
    while (r <= n) {
      if (every + get(2, 1, a[l] - 1) + every_suff > k) {
        upd(2, a[r], -1);
        every_suff -= get(2, 1, a[r] - 1);
        every -= get(1, a[r] + 1, 1e9);
        r++;
      } else
        break;
    }
    if (r > n) break;
    upd(1, a[l], 1);
    every += get(2, 1, a[l] - 1);
    res += n - r + 1;
  }
  printf("%I64d", res);
  return 0;
}
