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;
int r[400005];
int l[400005];
void update(int tree[], int node, int b, int e, int val, int op) {
  if (b > val || e < val) {
    return;
  }
  if (b <= val && val <= e) {
    tree[node] += op;
    if (b == e) {
      return;
    }
  }
  int md = (b + e) / 2;
  int left = 2 * node;
  int right = left + 1;
  update(tree, left, b, md, val, op);
  update(tree, right, md + 1, e, val, op);
  return;
}
void cnt(int tree[], int node, int b, int e, int x, int y, long long& t) {
  if (b > y || e < x) {
    return;
  }
  if (x <= b && e <= y) {
    t += tree[node];
    return;
  }
  int md = (b + e) / 2;
  int left = 2 * node;
  int right = left + 1;
  cnt(tree, left, b, md, x, y, t);
  cnt(tree, right, md + 1, e, x, y, t);
  return;
}
int main() {
  int n;
  long long k;
  scanf("%d %lld", &n, &k);
  pair<int, int> tmp[n];
  int ns[n];
  int t;
  for (int i = 0; i < n; i++) {
    scanf("%d", &t);
    tmp[i] = {t, i};
  }
  sort(tmp, tmp + n);
  int val = 0;
  for (int i = 0; i < n; i++) {
    if (i != 0 && tmp[i - 1].first == tmp[i].first) {
      ;
    } else {
      val++;
    }
    int pos = tmp[i].second;
    ns[pos] = val;
  }
  long long liv[n], riv[n];
  riv[n - 1] = 0;
  update(r, 1, 1, 100000, ns[n - 1], 1);
  for (int i = n - 2; i > 0; i--) {
    riv[i] = riv[i + 1];
    long long t = 0;
    cnt(r, 1, 1, 100000, 1, ns[i] - 1, t);
    riv[i] += t;
    update(r, 1, 1, 100000, ns[i], 1);
  }
  int i = 0, j = 1;
  long long ans = 0;
  long long xtra = 0;
  while (i != n - 1 && j != n) {
    long long large = 0;
    long long t = 0;
    cnt(l, 1, 1, 100000, ns[i] + 1, 100000, t);
    large = t;
    if (i != 0) {
      liv[i] = liv[i - 1] + large;
    } else {
      liv[i] = 0;
    }
    while (j != n) {
      long long small = 0;
      t = 0;
      cnt(r, 1, 1, 100000, 1, ns[i] - 1, t);
      small = t;
      long long inv = xtra + small;
      inv += riv[j];
      inv += liv[i];
      if (inv > k) {
        j++;
        update(r, 1, 1, 100000, ns[j - 1], -1);
        t = 0;
        cnt(l, 1, 1, 100000, ns[j - 1] + 1, 100000, t);
        xtra -= t;
      } else {
        ans += n - j;
        i++;
        xtra += small;
        update(l, 1, 1, 100000, ns[i - 1], 1);
        if (j == i) {
          j++;
          update(r, 1, 1, 100000, ns[j - 1], -1);
          t = 0;
          cnt(l, 1, 1, 100000, ns[j - 1] + 1, 100000, t);
          xtra -= t;
        }
        break;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}
