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;
int n, c[100010], f[100010], g[100010];
long long ans, sum;
int low[100010], high[100010];
struct A {
  int no, x;
} a[100010];
long long k;
inline bool cmp(const A &x1, const A &x2) { return x1.x < x2.x; }
inline void updata_f(int x, int y) {
  for (; x <= n; x += x & -x) f[x] += y;
}
inline int query_f(int x) {
  int tot = 0;
  for (; x; x -= x & -x) tot += f[x];
  return tot;
}
inline void updata_g(int x, int y) {
  for (; x <= n; x += x & -x) g[x] += y;
}
inline int query_g(int x) {
  int tot = 0;
  for (; x; x -= x & -x) tot += g[x];
  return tot;
}
int main() {
  scanf("%d%I64d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].x);
    a[i].no = i;
  }
  sort(a + 1, a + 1 + n, cmp);
  int cnt = 0, last = -1;
  for (int i = 1; i <= n; i++) {
    if (last != a[i].x) {
      cnt++;
      last = a[i].x;
    }
    c[a[i].no] = cnt;
  }
  for (int i = n; i >= 1; i--) {
    updata_g(c[i], 1);
    sum += query_g(c[i] - 1);
  }
  int r = 2;
  updata_f(c[1], 1);
  updata_g(c[1], -1);
  for (int i = 1; i <= n; i++) {
    if (i == r)
      sum -= i - query_f(c[r]) + query_g(c[r] - 1), updata_g(c[r], -1), r++;
    while (r <= n && sum > k) {
      sum -= i - query_f(c[r]) + query_g(c[r] - 1);
      updata_g(c[r], -1);
      r++;
    }
    ans += n + 1 - r;
    if (i < n) {
      updata_f(c[i + 1], 1);
      sum += i + 1 - query_f(c[i + 1]) + query_g(c[i + 1] - 1);
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
