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>
const int INF = 0x3f3f3f3f;
long long int getint() {
  long long int ans = 0;
  int f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') f = -f;
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return ans * f;
}
int lowbit(const int &a) { return ((a) & (-a)); }
using namespace std;
const int MAXN = 100000;
int hsh[MAXN + 10], a[MAXN + 10], cnt;
int L[MAXN + 10], R[MAXN + 10];
bool cmp(const int &i, const int &j) { return a[i] < a[j]; }
int sum(const int c[], int pos) {
  int ret = 0;
  while (pos > 0) {
    ret += c[pos];
    pos -= lowbit(pos);
  }
  return ret;
}
void update(int c[], int pos, int num) {
  while (pos <= cnt) {
    c[pos] += num;
    pos += lowbit(pos);
  }
}
int main() {
  int n = getint();
  long long int ans = 0;
  long long int k = getint();
  for (int i = 1; i <= n; ++i) {
    a[i] = getint();
    hsh[i] = i;
  }
  sort(hsh + 1, hsh + n + 1, cmp);
  cnt = 1;
  for (int i = 1; i <= n; ++i, ++cnt) {
    while (i < n && a[hsh[i]] == a[hsh[i + 1]]) a[hsh[i]] = cnt, ++i;
    a[hsh[i]] = cnt;
  }
  --cnt;
  int l = 1, r = 2;
  long long int num = 0;
  update(L, a[l], 1);
  for (int i = 2; i <= n; ++i) {
    update(R, a[i], 1);
    num += a[i] < a[1];
    num += sum(R, cnt) - sum(R, a[i]);
  }
  ans = (num > k);
  while (l < n) {
    if (l == r) {
      update(R, a[r], -1);
      num -= sum(R, a[r] - 1);
      num -= sum(L, cnt) - sum(L, a[r]);
      ++r;
      if (num > k) ++ans;
    }
    while (r < n) {
      long long int q = sum(R, a[r] - 1) + sum(L, cnt) - sum(L, a[r]);
      if (num - q <= k) break;
      num -= q;
      ++ans;
      update(R, a[r], -1);
      ++r;
    }
    ++l;
    update(L, a[l], 1);
    num += sum(R, a[l] - 1);
    num += sum(L, cnt) - sum(L, a[l]);
    ans += r - l;
  }
  printf("%I64d\n", (long long int)n * (n - 1) / 2 - ans);
  return 0;
}
