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 = 100005, PRE = 0, SUF = 1;
struct node {
  int key, id;
  node(int key_ = 0, int id_ = 0) {
    key = key_;
    id = id_;
  }
} a_[MAXN];
int n, cnt, a[MAXN], t[2][MAXN];
long long k;
bool cmp(node x, node y) {
  return (x.key == y.key) ? (x.id < y.id) : (x.key < y.key);
}
int lowbit(int x) { return x & (-x); }
void add(int o, int x, int y) {
  while (x <= MAXN) {
    t[o][x] = t[o][x] + y;
    x = x + lowbit(x);
  }
}
int query(int o, int x) {
  int re = 0;
  while (x) {
    re = re + t[o][x];
    x = x - lowbit(x);
  }
  return re;
}
int main() {
  scanf("%d%lld", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a_[i].key);
    a_[i].id = i;
  }
  sort(a_ + 1, a_ + 1 + n, cmp);
  for (int i = 1; i <= n; i++) {
    if (a_[i].key != a_[i - 1].key) cnt++;
    a[a_[i].id] = cnt;
  }
  long long now = 0;
  for (int i = 1; i <= n; i++) {
    now = now + ((i - 1) - query(PRE, a[i]));
    add(PRE, a[i], 1);
  }
  if (now <= k) {
    printf("%lld", (long long)n * (n - 1) / 2);
    return 0;
  }
  memset(t[PRE], 0, sizeof(t[PRE]));
  for (int i = n; i >= 1; i--) add(SUF, a[i], 1);
  add(PRE, a[1], 1);
  add(SUF, a[1], -1);
  int r = 2;
  long long ans = 0;
  for (int l = 1; l <= n - 1 && r <= n; l++) {
    while (now > k && r <= n) {
      now = now - (l - query(PRE, a[r]));
      now = now - (query(SUF, a[r] - 1));
      add(SUF, a[r], -1);
      r++;
    }
    if (r <= n)
      ans = ans + (n - r + 1);
    else
      break;
    now = now + (l - query(PRE, a[l + 1]));
    now = now + (query(SUF, a[l + 1] - 1));
    add(PRE, a[l + 1], 1);
  }
  printf("%lld", ans);
}
