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 gcd(int a, int b) {
  return (b > a) ? gcd(b, a) : (b == 0) ? a : gcd(b, a % b);
}
int bitcount(int a) { return (a != 0) ? bitcount(a & (a - 1)) + 1 : 0; }
long long bigmod(long long num, long long pow, long long prime) {
  if (pow == 0ll) return 1ll;
  long long ans = bigmod(num, pow / 2ll, prime);
  ans = (ans * ans) % prime;
  if (pow % 2ll) ans = (ans * num) % prime;
  return ans;
}
long long N, K, a[100005], b[100005], c[100005];
long long d[100005], e[100005];
map<long long, long long> m;
void acompress() {
  for (int i = (0); i < (N); ++i) b[i] = a[i];
  sort(b, b + N);
  for (int i = (0); i < (N); ++i) {
    if (m.find(b[i]) == m.end()) m[b[i]] = i + 1;
  }
  for (int i = (0); i < (N); ++i) a[i] = m[a[i]];
  return;
}
void upd(long long dd[100005], long long pp, long long vl) {
  while (pp <= 100001) {
    dd[pp] += vl;
    pp += pp & -pp;
  }
  return;
}
long long qry(long long dd[100005], long long pp) {
  long long ans = 0;
  while (pp > 0) {
    ans += dd[pp];
    pp -= pp & -pp;
  }
  return ans;
}
int main() {
  long long sum, ans;
  cin >> N >> K;
  for (int i = (0); i < (N); ++i) cin >> a[i];
  acompress();
  sum = ans = 0;
  for (int i = (0); i < (N); ++i) {
    sum += i - qry(d, a[i]);
    upd(d, a[i], 1);
  }
  for (long long i = N - 1, j = N - 1; i > 0; i--) {
    sum += qry(e, a[i] - 1);
    sum += j + 1 - qry(d, a[i]);
    upd(e, a[i], 1);
    while (j >= i || sum > K) {
      if (j < 0) break;
      upd(d, a[j], -1);
      long long now = qry(e, a[j] - 1) + j - qry(d, a[j]);
      sum -= now;
      j--;
    }
    ans += j + 1;
  }
  cout << ans << endl;
  return 0;
}
