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;
template <class T>
vector<int> compress(T first, T last) {
  vector<typename iterator_traits<T>::value_type> cv(first, last);
  vector<int> ret(cv.size());
  sort(cv.begin(), cv.end());
  cv.erase(unique(cv.begin(), cv.end()), cv.end());
  for (int i = 0; first != last; first++, i++)
    ret[i] = lower_bound(cv.begin(), cv.end(), *first) - cv.begin();
  return ret;
}
struct BIT {
  vector<long long> dat;
  BIT(int n) : dat(n) {}
  void update(int k, long long x) {
    k++;
    for (; k < dat.size(); k += k & -k) dat[k] += x;
  }
  long long sumFront(int k) {
    k++;
    long long ret = 0;
    for (; k > 0; k -= k & -k) ret += dat[k];
    return ret;
  }
  long long sumBack(int k) {
    return sumFront(dat.size() - 2) - sumFront(k - 1);
  }
};
int n;
long long K;
vector<int> a;
long long inv;
BIT bitL(101010);
BIT bitR(101010);
void addLeft(int k) {
  inv += bitL.sumBack(a[k] + 1);
  inv += bitR.sumFront(a[k] - 1);
  bitL.update(a[k], 1);
}
void removeRight(int k) {
  inv -= bitL.sumBack(a[k] + 1);
  inv -= bitR.sumFront(a[k] - 1);
  bitR.update(a[k], -1);
}
int main() {
  cin >> n >> K;
  a.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  a = compress(a.begin(), a.end());
  for (int i = 0; i < n; i++) {
    inv += bitR.sumBack(a[i] + 1);
    bitR.update(a[i], 1);
  }
  long long ans = 0;
  int j = 0;
  addLeft(0);
  for (int i = 1; i < n; i++) {
    while (j < n && j < i) removeRight(j++);
    while (j < n && inv > K) removeRight(j++);
    ans += n - j;
    addLeft(i);
  }
  cout << ans << endl;
}
