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 = 1e5 + 5;
struct st {
  int x, id;
} lsh[N];
int n, t[N], tot, a[N], T[N];
long long k, ans;
bool cmp(st a, st b) { return a.x < b.x; }
void add(int x, int y) {
  for (; x <= n; x += x & -x) t[x] += y;
}
void addT(int x, int y) {
  for (; x <= n; x += x & -x) T[x] += y;
}
int ask(int x) {
  int res = 0;
  for (; x; x -= x & -x) res += t[x];
  return res;
}
int askT(int x) {
  int res = 0;
  for (; x; x -= x & -x) res += T[x];
  return res;
}
int main() {
  scanf("%d%lld", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &lsh[i].x);
    lsh[i].id = i;
  }
  sort(lsh + 1, lsh + 1 + n, cmp);
  for (int i = 1; i <= n; i++) {
    if (lsh[i].x != lsh[i - 1].x) ++tot;
    a[lsh[i].id] = tot;
  }
  long long now = 0;
  for (int i = 1; i <= n; i++) {
    now += i - 1 - ask(a[i]);
    add(a[i], 1);
  }
  int r = 2;
  add(a[1], -1);
  now -= ask(a[1] - 1);
  for (int i = 1; i <= n; i++) {
    now += i - 1 - askT(a[i]);
    now += ask(a[i] - 1);
    addT(a[i], 1);
    while (now > k && r <= n) {
      add(a[r], -1);
      now -= i - askT(a[r]);
      now -= ask(a[r] - 1);
      r++;
    }
    if (r > n) break;
    if (i == r) {
      add(a[r], -1);
      now -= i - askT(a[r]);
      now -= ask(a[r] - 1);
      r++;
    }
    ans += n - r + 1;
  }
  printf("%lld", ans);
  return 0;
}
