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 = 100010;
int n, cnt;
long long k;
int a[maxn], b[maxn];
struct BIT {
  int sum[maxn];
  void clear() {
    for (int i = 0; i <= cnt; i++) sum[i] = 0;
  }
  void add(int i, int a) {
    for (; i <= cnt; i += (i) & (-i)) sum[i] += a;
  }
  int ask(int i) {
    int ans = 0;
    for (; i > 0; i -= (i) & (-i)) ans += sum[i];
    return ans;
  }
} b1, b2;
int main() {
  while (cin >> n >> k) {
    for (int i = 0; i < n; i++) scanf("%d", a + i);
    for (int i = 0; i < n; i++) b[i] = a[i];
    sort(b, b + n);
    cnt = unique(b, b + n) - b;
    for (int i = 0; i < n; i++) a[i] = lower_bound(b, b + cnt, a[i]) - b + 1;
    cnt += 4;
    b1.clear();
    b2.clear();
    long long now = 0, ans = 0;
    for (int i = 0; i < n; i++) {
      now += b1.ask(cnt - a[i] - 1);
      b1.add(cnt - a[i], 1);
    }
    for (int i = n - 1, j = n - 1; i >= 0; i--) {
      while (i < j) {
        int l = b1.ask(cnt - a[j] - 1);
        int r = b2.ask(a[j] - 1);
        if (now + l + r > k) break;
        now += l + r;
        b2.add(a[j], 1);
        j--;
      }
      ans += n - j - 1;
      now -= b2.ask(a[i] - 1);
      now -= b1.ask(cnt - a[i] - 1);
      b1.add(cnt - a[i], -1);
    }
    cout << ans << endl;
  }
  return 0;
}
