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 long double pi = 3.14159265359;
template <typename T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
void chmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void chmax(T &x, T y) {
  x = max(x, y);
}
const int maxn = 1e5 + 5;
struct tree {
  int t[maxn * 4];
  tree() { fill(t, t + maxn * 4, 0); }
  void modify(int v, int l, int r, int i, int x) {
    if (l == r) {
      t[v] += x;
      return;
    }
    t[v] += x;
    int c = (l + r) >> 1;
    if (i <= c) {
      modify(2 * v, l, c, i, x);
    } else {
      modify(2 * v + 1, c + 1, r, i, x);
    }
  }
  int sum(int v, int l, int r, int i, int j) {
    if (i <= l && r <= j) {
      return t[v];
    }
    int res = 0;
    int c = (l + r) >> 1;
    if (i <= c) {
      res += sum(2 * v, l, c, i, j);
    }
    if (c < j) {
      res += sum(2 * v + 1, c + 1, r, i, j);
    }
    return res;
  }
};
tree tl, tr;
int main() {
  srand(time(NULL));
  int n;
  scanf("%d", &n);
  long long k;
  cin >> k;
  vector<int> a, cc;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    a.push_back(x);
    cc.push_back(x);
  }
  sort(cc.begin(), cc.end());
  cc.resize(unique(cc.begin(), cc.end()) - cc.begin());
  for (int &x : a) {
    x = lower_bound(cc.begin(), cc.end(), x) - cc.begin() + 1;
  }
  long long ans = 0;
  int r = n;
  long long s = 0;
  for (int i = 0; i < n; i++) {
    if (i == r) {
      s -= tl.sum(1, 0, maxn - 1, a[i] + 1, maxn - 1);
      tr.modify(1, 0, maxn - 1, a[i], -1);
      s -= tr.sum(1, 0, maxn - 1, 0, a[i] - 1);
      r++;
    }
    s += tl.sum(1, 0, maxn - 1, a[i] + 1, maxn - 1);
    s += tr.sum(1, 0, maxn - 1, 0, a[i] - 1);
    tl.modify(1, 0, maxn - 1, a[i], 1);
    while (i < r - 1 && s <= k) {
      r--;
      s += tr.sum(1, 0, maxn - 1, 0, a[r] - 1);
      s += tl.sum(1, 0, maxn - 1, a[r] + 1, maxn - 1);
      tr.modify(1, 0, maxn - 1, a[r], 1);
    }
    while (r < n && s > k) {
      s -= tl.sum(1, 0, maxn - 1, a[r] + 1, maxn - 1);
      tr.modify(1, 0, maxn - 1, a[r], -1);
      s -= tr.sum(1, 0, maxn - 1, 0, a[r] - 1);
      r++;
    }
    if (s <= k) {
      ans += n - r;
    }
  }
  cout << ans << endl;
  return 0;
}
