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 N = 1e5 + 10;
inline long long in() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
long long n, arr[N], treeL[N], treeR[N];
long long k;
map<long long, long long> m;
vector<long long> val;
vector<long long>::iterator it;
long long c1(long long x) {
  long long sum = 0;
  for (; x > 0; x -= x & (-x)) sum += treeL[x];
  return sum;
}
void update1(long long i, long long val) {
  for (; i <= n; i += i & (-i)) treeL[i] += val;
}
long long c2(long long x) {
  long long sum = 0;
  for (; x > 0; x -= x & (-x)) sum += treeR[x];
  return sum;
}
void update2(long long i, long long val) {
  for (; i <= n; i += i & (-i)) treeR[i] += val;
}
int main(int argc, const char* argv[]) {
  n = in();
  k = in();
  val.push_back(-1);
  for (long long i = 1; i <= n; i++) {
    long long a = in();
    arr[i] = a;
    val.push_back(a);
  }
  sort(val.begin(), val.end());
  it = unique(val.begin(), val.end());
  val.resize(distance(val.begin(), it));
  for (long long i = 1; i < val.size(); i++) {
    m[val[i]] = i;
  }
  for (long long i = 1; i <= n; i++) {
    arr[i] = m[arr[i]];
  }
  long long inv = 0, r = 2;
  for (long long i = n; i >= 2; i--) {
    long long L = arr[i] - 1;
    inv += c2(L);
    update2(arr[i], 1);
  }
  long long ans = 0;
  for (long long i = 1; i <= n - 1; i++) {
    long long L = arr[i] - 1;
    if (i == r) {
      update2(arr[r], -1);
      inv -= c2(arr[r] - 1);
      inv -= c1(n) - c1(arr[r]);
      r++;
    }
    inv += c2(L) + c1(n) - c1(L + 1);
    update1(arr[i], 1);
    while (inv > k && r <= n) {
      update2(arr[r], -1);
      inv -= c2(arr[r] - 1);
      inv -= c1(n) - c1(arr[r]);
      r++;
    }
    if (r <= n && i < r) ans += n - r + 1;
  }
  cout << ans;
  return 0;
}
