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 <typename T>
T next_int() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') {
    p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * p;
}
const int max_n = (int)1e6 + 227 + 1;
const int max_int = (int)1e9 + 227 + 1;
const int mod = (int)1e9 + 7;
int d[2][max_n];
void inc(int type, int p, int a) {
  for (int i = p; i < max_n; i += (i & (-i))) {
    d[type][i] += a;
  }
}
int pre(int type, int p) {
  int ans = 0;
  for (int i = p; i > 0; i -= (i & (-i))) {
    ans += d[type][i];
  }
  return ans;
}
int suf(int type, int p) { return pre(type, max_n - 1) - pre(type, p - 1); }
long long check(int a, long long mda) {
  mda += suf(0, a + 1);
  mda += pre(1, a - 1);
  return mda;
}
int main() {
  int n;
  cin >> n;
  long long m;
  cin >> m;
  vector<int> a(n);
  vector<pair<int, int> > b(n);
  for (int i = 0; i < n; i++) {
    a[i] = next_int<int>();
    b[i] = make_pair(a[i], i);
  }
  sort(b.begin(), b.end());
  int k = 0;
  for (int i = 0; i < n; i++) {
    if (!i || b[i].first != b[i - 1].first) k++;
    a[b[i].second] = k;
  }
  long long mda = 0;
  for (int i = 0; i < n; i++) {
    inc(0, a[i], 1);
    mda += suf(0, a[i] + 1);
  }
  long long ans = 0;
  if (mda <= m) ans += n - 1;
  int r = n - 1;
  for (int l = n - 1; l >= 1; l--) {
    inc(0, a[l], -1);
    mda -= suf(0, a[l] + 1);
    mda -= pre(1, a[l] - 1);
    if (l == n - 1) continue;
    while (l <= (r - 1) && check(a[r], mda) <= m) {
      mda += suf(0, a[r] + 1);
      mda += pre(1, a[r] - 1);
      inc(1, a[r], 1);
      r--;
    }
    ans += n - r - 1;
  }
  cout << ans << "\n";
}
