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;
struct Bit {
  int a[1020304];
  void clr() { memset(a, 0, sizeof(a)); }
  void Ins(int t, int k) {
    for (int z = t; z < 1020304; z += (z & (-z))) a[z] += k;
  }
  int ask(int p) {
    if (p == 0) return 0;
    int ret = 0;
    for (int z = p; z; z -= (z & (-z))) ret += a[z];
    return ret;
  }
  int Ask(int l, int r) { return ask(r) - ask(l - 1); }
} t1, t2;
int n, a[1020304], b[1020304];
long long K;
map<int, int> MAP;
int main() {
  cin >> n >> K;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), b[i] = a[i];
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; ++i) MAP[b[i]] = i;
  for (int i = 1; i <= n; ++i) a[i] = MAP[a[i]];
  t2.clr();
  t1.clr();
  long long tNow = 0;
  for (int i = n; i >= 2; --i) {
    tNow += t2.Ask(1, a[i] - 1);
    t2.Ins(a[i], 1);
  }
  int tJ = 0;
  long long ans = 0;
  for (int i = 1; i <= n - 1; ++i) {
    while (tNow + t1.Ask(a[tJ + 1] + 1, n) + t2.Ask(1, a[tJ + 1] - 1) <= K &&
           tJ < i) {
      tNow += t1.Ask(a[tJ + 1] + 1, n) + t2.Ask(1, a[tJ + 1] - 1);
      t1.Ins(a[++tJ], 1);
    }
    ans += tJ;
    tNow -= t1.Ask(a[i + 1] + 1, n) + t2.Ask(1, a[i + 1] - 1);
    t2.Ins(a[i + 1], -1);
  }
  cout << ans << endl;
  return 0;
}
