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;
void read(int &x) {
  char ch = getchar();
  int f = 1;
  while ((ch != '-') && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-')
    f = -1, x = 0;
  else
    x = ch - 48;
  ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  x *= f;
}
void read(long long &x) {
  char ch = getchar();
  int f = 1;
  while ((ch != '-') && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-')
    f = -1, x = 0;
  else
    x = ch - 48;
  ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10LL + ch - 48, ch = getchar();
  x *= f;
}
void _print(const int &x) {
  if (x < 10)
    putchar(48 + x);
  else
    _print(x / 10), _print(x % 10);
}
void _print(const long long &x) {
  if (x < 10)
    putchar(48 + x);
  else
    _print(x / 10), _print(x % 10);
}
void print(const int &x) {
  if (x < 0)
    putchar('-'), _print(-x);
  else
    _print(x);
}
void print(const long long &x) {
  if (x < 0)
    putchar('-'), _print(-x);
  else
    _print(x);
}
void println(const int &x) {
  print(x);
  putchar('\n');
}
void println(const long long &x) {
  print(x);
  putchar('\n');
}
int n, a[100010], bin[100010];
long long K;
int cnt;
int b[100010] = {0}, c[100010] = {0};
long long liv[100010], riv[100010];
void add(int *A, int x, int d) {
  for (; x <= cnt; x += x & -x) A[x] += d;
}
long long ask(int *A, int x) {
  long long ret = 0;
  for (; x; x -= x & -x) ret += A[x];
  return ret;
}
int main() {
  read(n);
  read(K);
  for (int i = (1); i <= (n); i++) read(a[i]), bin[i] = a[i];
  sort(bin + 1, bin + n + 1);
  cnt = unique(bin + 1, bin + n + 1) - bin - 1;
  for (int i = (1); i <= (n); i++) {
    a[i] = lower_bound(bin + 1, bin + cnt + 1, a[i]) - bin;
    add(b, a[i], 1);
    liv[i] = liv[i - 1] + ask(b, cnt) - ask(b, a[i]);
  }
  memset(b, (0), sizeof(b));
  for (int i = (n); i >= (1); i--) {
    add(b, a[i], 1);
    riv[i] = riv[i + 1] + ask(b, a[i] - 1);
  }
  int r = n;
  while (r >= 1 && riv[r] <= K) r--;
  r++;
  memset(b, (0), sizeof(b));
  for (int i = (r); i <= (n); i++) add(b, a[i], 1);
  long long ans = 0, tmp = 0;
  for (int l = (1); l <= (n - 1); l++) {
    add(c, a[l], 1);
    tmp += ask(b, a[l] - 1);
    while (l >= r) {
      tmp -= ask(c, cnt) - ask(c, a[r]);
      add(b, a[r], -1);
      r++;
    }
    while (r <= n && liv[l] + riv[r] + tmp > K) {
      tmp -= ask(c, cnt) - ask(c, a[r]);
      add(b, a[r], -1);
      r++;
    }
    ans += n - r + 1;
  }
  cout << ans << endl;
  return 0;
}
