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 MAX = 100 * 1000;
const int INF = 1 << 29;
const long long D = 1000 * 1000 * 1000 + 7;
long long inv[2][MAX + 10];
int a[MAX + 10];
int b[MAX + 10];
int head;
int n, k;
long long Inv;
long long cnt;
int t[2][MAX + 10];
void update(bool, int, int);
int sum(bool, int);
void clear();
void read();
void init();
void add(int);
void del(int);
int p1, p2;
int main() {
  read();
  init();
  p2 = 1;
  for (int i = 0; i < n - 1 && p2 < n; i++) {
    if (p2 == i) del(p2++);
    p1 = i;
    add(i);
    while (p2 < n && Inv > k) del(p2++);
    cnt += n - p2;
  }
  cout << cnt << endl;
  return 0;
}
void add(int i) {
  update(0, (lower_bound(b, b + head, a[i]) - b) + 1, 1);
  Inv += sum(1, (lower_bound(b, b + head, a[i]) - b));
  Inv += (p1 + 1) - sum(0, (lower_bound(b, b + head, a[i]) - b) + 1);
}
void del(int i) {
  update(1, (lower_bound(b, b + head, a[i]) - b) + 1, -1);
  Inv -= sum(1, (lower_bound(b, b + head, a[i]) - b));
  Inv -= (p1 + 1) - sum(0, (lower_bound(b, b + head, a[i]) - b) + 1);
}
void read() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i], b[i] = a[i];
}
void clear(bool f) {
  for (int i = 0; i <= n; i++) t[f][i] = 0;
}
void update(bool f, int x, int val) {
  for (; x <= n; x += x & -x) t[f][x] += val;
}
int sum(bool f, int x) {
  int ret = 0;
  for (; x > 0; x -= x & -x) ret += t[f][x];
  return ret;
}
void init() {
  sort(b, b + n);
  head = unique(b, b + n) - b;
  for (int i = n - 1; i >= 1; i--) {
    Inv += sum(1, (lower_bound(b, b + head, a[i]) - b));
    update(1, (lower_bound(b, b + head, a[i]) - b) + 1, 1);
  }
}
