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 a[110000], b[110000];
int n;
long long k;
struct IndexArray {
  int s[110000 + 5];
  int N;
  void init(int n) {
    N = n;
    fill(s, s + n + 1, 0);
  }
  int lowbit(int t) { return t & (-t); }
  void update(int i, int val) {
    while (i <= N) {
      s[i] += val;
      i += lowbit(i);
    }
  }
  int get_sum(int i) {
    int ret = 0;
    while (i > 0) {
      ret += s[i];
      i -= lowbit(i);
    }
    return ret;
  }
} bit1, bit2;
int force() {
  int ans = 0;
  for (int i = 1; i < n; i++)
    for (int j = i + 1; j <= n; j++) {
      vector<int> v;
      for (int p = 1; p <= i; p++) v.push_back(a[p]);
      for (int p = j; p <= n; p++) v.push_back(a[p]);
      int sum = 0;
      for (int i = 0; i < (int)v.size(); i++)
        for (int j = 0; j < i; j++)
          if (v[i] < v[j]) sum++;
      if (sum <= k) ans++;
    }
  printf("ans=%d\n", ans);
  return ans;
}
int main() {
  while (cin >> n >> k) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
      b[i] = a[i];
    }
    sort(b + 1, b + 1 + n);
    int x = unique(b + 1, b + 1 + n) - b;
    for (int i = 1; i <= n; i++) {
      a[i] = lower_bound(b + 1, b + x, a[i]) - b;
    }
    bit1.init(n);
    bit2.init(n);
    long long ans = 0;
    long long inv1 = 0, inv2 = 0, inv3 = 0;
    bit1.update(a[1], 1);
    int j;
    for (j = n; j >= 2; j--) {
      int s2 = bit2.get_sum(a[j] - 1);
      int s3 = 1 - bit1.get_sum(a[j]);
      if (inv1 + inv2 + inv3 + s2 + s3 <= k) {
        bit2.update(a[j], 1);
        inv2 += s2;
        inv3 += s3;
      } else {
        break;
      }
    }
    j++;
    if (j == 2) {
      cout << n * (long long)(n - 1) / 2 << endl;
      continue;
    }
    ans += n - j + 1;
    for (int i = 2; i < n; i++) {
      inv1 += i - 1 - bit1.get_sum(a[i]);
      bit1.update(a[i], 1);
      inv3 += bit2.get_sum(a[i] - 1);
      while ((inv1 + inv2 + inv3 > k) && j <= n) {
        bit2.update(a[j], -1);
        inv2 -= bit2.get_sum(a[j] - 1);
        inv3 -= i - bit1.get_sum(a[j]);
        j++;
      }
      ans += n - j + 1;
    }
    cout << ans << endl;
  }
}
