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 N = 1e5 + 10;
struct cntSet {
 public:
  void erase(int);
  void insert(int);
  int getLower(int);
  int getHigher(int);

 private:
  int fen[N];
  int getSum(int);
  void add(int, int);
};
int n, a[N], db[N], dbs;
cntSet _right, _left;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    db[i] = a[i];
  }
  sort(db, db + n);
  dbs = unique(db, db + n) - db;
  for (int i = 0; i < n; i++) a[i] = lower_bound(db, db + dbs, a[i]) - db;
  long long curInv = 0;
  for (int i = n - 1; i > 0; i--) {
    curInv += _right.getLower(a[i]);
    _right.insert(a[i]);
  }
  long long ans = 0;
  curInv += _right.getLower(a[0]);
  if (curInv <= k) ans += n - 1;
  curInv -= _right.getLower(a[0]);
  int r = 1;
  for (int l = 0; l < n; l++) {
    bool flag = true;
    while ((flag && curInv > k) || r == l)
      if (r < n) {
        _right.erase(a[r]);
        curInv -= _left.getHigher(a[r]) + _right.getLower(a[r]);
        r++;
      } else
        flag = false;
    if (!flag) break;
    if (l > 0) ans += n - r;
    _left.insert(a[l]);
    curInv += _left.getHigher(a[l]) + _right.getLower(a[l]);
  }
  cout << ans << endl;
}
void cntSet::insert(int v) { add(v, +1); }
void cntSet::erase(int v) { add(v, -1); }
int cntSet::getLower(int v) { return getSum(v); }
int cntSet::getHigher(int v) { return getSum(dbs) - getSum(v + 1); }
void cntSet::add(int i, int val) {
  for (i++; i <= dbs; i += i & -i) fen[i] += val;
}
int cntSet::getSum(int r) {
  int res = 0;
  for (; r; r ^= r & -r) res += fen[r];
  return res;
}
