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 MAXN = 1e5 + 10;
long long k;
long long curk = 0, ans = 0;
int n, c[2][MAXN], a[MAXN], N;
void update(int k, int pos, int val) {
  while (pos <= N) {
    c[k][pos] += val;
    pos += pos & -pos;
  }
}
int query(int k, int pos) {
  int ret = 0;
  while (pos > 0) {
    ret += c[k][pos];
    pos -= pos & -pos;
  }
  return ret;
}
int main() {
  scanf("%d%I64d", &n, &k);
  vector<int> v;
  map<int, int> num;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), v.push_back(a[i]);
  N = 0;
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++)
    if (i == 0 || v[i] != v[i - 1]) num[v[i]] = ++N;
  for (int i = 1; i <= n; i++) a[i] = num[a[i]];
  memset(c, 0, sizeof(c));
  long long temp = 0;
  int r;
  for (int i = n; i >= 1; i--) {
    temp += query(1, a[i] - 1);
    update(1, a[i], 1);
    if (temp <= k) {
      curk = temp;
      r = i;
    } else {
      update(1, a[i], -1);
      break;
    }
  }
  for (int i = 1; i <= n; i++) {
    while (r <= i) {
      curk -= query(1, a[r] - 1);
      curk -= query(0, N) - query(0, a[r]);
      update(1, a[r], -1);
      r++;
    }
    if (r == n + 1) break;
    curk += query(0, N) - query(0, a[i]);
    curk += query(1, a[i] - 1);
    update(0, a[i], 1);
    while (r <= n && curk > k) {
      curk -= query(1, a[r] - 1);
      curk -= query(0, N) - query(0, a[r]);
      update(1, a[r], -1);
      r++;
    }
    if (r == n + 1) break;
    ans += n - r + 1;
  }
  cout << ans << endl;
}
