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;
#pragma comment(linker, "/STACK:200000000")
const double EPS = 1E-9;
const int INF = 1000000000;
const long long INF64 = (long long)1E18;
const double PI = 3.1415926535897932384626433832795;
int a[110000], m;
long long pref[110000], suf[110000];
struct tree {
  vector<int> t;
  tree() { t.assign(m, 0); }
  int count(int i) {
    int res = 0;
    for (; i >= 0; i = (i & (i + 1)) - 1) res += t[i];
    return res;
  }
  void add(int i, int d) {
    for (; i < m; i = (i | (i + 1))) t[i] += d;
  }
  int sum(int l, int r) {
    if (l > r) return 0;
    int res = count(r);
    if (l) res -= count(l - 1);
    return res;
  }
};
int main() {
  int n;
  long long k;
  cin >> n >> k;
  vector<int> v;
  for (int i = 0; i < (int)(n); i++) {
    scanf("%d", &a[i]);
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  m = (int)v.size();
  for (int i = 0; i < (int)(n); i++)
    a[i] = int(lower_bound(v.begin(), v.end(), a[i]) - v.begin());
  tree t1;
  for (int i = 0; i < (int)(n); i++) {
    if (i)
      pref[i] = pref[i - 1] + t1.sum(a[i] + 1, m - 1);
    else
      pref[i] = 0;
    t1.add(a[i], +1);
  }
  tree t2;
  for (int i = (int)(n)-1; i >= 0; i--) {
    if (i < n - 1)
      suf[i] = suf[i + 1] + t2.sum(0, a[i] - 1);
    else
      suf[i] = 0;
    t2.add(a[i], +1);
  }
  tree left, right;
  int r = 1;
  for (int i = 1; i < n; i++) right.add(a[i], +1);
  long long lst = 0;
  long long ans = 0;
  for (int l = 0; l < (int)(n); l++) {
    while (l >= r) {
      lst -= left.sum(a[r] + 1, m - 1);
      right.add(a[r], -1);
      r++;
    }
    left.add(a[l], +1);
    lst += right.sum(0, a[l] - 1);
    while (r < n && pref[l] + suf[r] + lst > k) {
      lst -= left.sum(a[r] + 1, m - 1);
      right.add(a[r], -1);
      r++;
    }
    ans += n - r;
  }
  cout << ans << endl;
  return 0;
}
