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 maxn = 1e5 + 10;
struct BIT {
  int bit[maxn];
  void upd(int x, int v) {
    for (; x < maxn; x += (x & -x)) bit[x] += v;
  }
  int soma(int x) {
    int ans = 0;
    for (; x > 0; x -= (x & -x)) ans += bit[x];
    return ans;
  }
} b_pref, b_suf;
int n;
long long k;
int a[maxn];
void compress(void) {
  map<int, int> mp;
  for (int i = 1; i <= n; i++) mp[a[i]] = 0;
  int aux = 0;
  for (auto &x : mp) x.second = ++aux;
  for (int i = 1; i <= n; i++) a[i] = mp[a[i]];
}
int main(void) {
  scanf("%d %lld", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  compress();
  long long inv = 0;
  for (int i = n; i >= 2; i--) {
    inv += 1ll * b_suf.soma(a[i] - 1);
    b_suf.upd(a[i], 1);
  }
  long long ans = 0;
  int l = 1, r = 2;
  for (; l <= n; l++) {
    inv += 1ll * (b_pref.soma(n) - b_pref.soma(a[l]));
    b_pref.upd(a[l], 1);
    inv += 1ll * b_suf.soma(a[l] - 1);
    while ((inv > k || r <= l) && r <= n) {
      inv -= 1ll * b_suf.soma(a[r] - 1);
      inv -= 1ll * (b_pref.soma(n) - b_pref.soma(a[r]));
      b_suf.upd(a[r], -1);
      ++r;
    }
    if (r <= n) ans += 1ll * (n - r + 1);
  }
  printf("%lld\n", ans);
}
