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;
long long n, k;
const int maxn = 1e5 + 5;
int a[maxn], ind[maxn];
long long pre[maxn], suf[maxn];
void compress() {
  vector<pair<int, int> > vec;
  for (int i = 0; i < n; i++) {
    vec.push_back({a[i], i});
  }
  sort(vec.begin(), vec.end());
  int c = 0;
  for (int i = 0; i < n; i++) {
    int x = vec[i].first;
    while (i < n && vec[i].first == x) {
      ind[vec[i].second] = c;
      i++;
    }
    c++;
    i--;
  }
}
void add2(int x, int val) {
  x++;
  for (; x < maxn; x += (x & -x)) {
    pre[x] += val;
  }
}
void add1(int x, int val) {
  x++;
  for (; x < maxn; x += (x & -x)) {
    suf[x] += val;
  }
}
long long get2(int x) {
  long long res = 0;
  for (; x; x &= x - 1) {
    res += pre[x];
  }
  return res;
}
long long get2(int l, int r) { return get2(r) - get2(l); }
long long get1(int x) {
  long long res = 0;
  for (; x; x &= x - 1) {
    res += suf[x];
  }
  return res;
}
long long get1(int l, int r) { return get1(r) - get1(l); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  compress();
  long long cnt = 0;
  for (int i = n - 1; i >= 0; i--) {
    cnt += get1(0, ind[i]);
    add1(ind[i], 1);
  }
  int r = 0;
  long long ans = 0;
  for (int l = 0; l < n; l++) {
    add2(ind[l], 1);
    cnt += get2(ind[l] + 1, n + 1) + get1(0, ind[l]);
    while (r < n && (r <= l || cnt > k)) {
      cnt -= get1(0, ind[r]) + get2(ind[r] + 1, n + 1);
      add1(ind[r], -1);
      r++;
    }
    if (cnt > k) {
      break;
    }
    ans += (n - r);
  }
  cout << ans;
  return 0;
}
