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 siz = 0;
const int MAXN = 1e5 + 100;
int n;
long long fen[MAXN][2], pref[MAXN], suff[MAXN];
int f[MAXN], F[MAXN], a[MAXN];
void add(int p, int k, int x) {
  p++;
  for (p; p <= n; p += (p & -p)) {
    fen[p][k] += x;
  }
  return;
}
long long ask(int p, int k) {
  p++;
  long long res = 0;
  for (p; p > 0; p -= (p & -p)) {
    res += fen[p][k];
  }
  return res;
}
void cler() {
  for (int i = 0; i <= n; i++) {
    fen[i][0] = 0;
    fen[i][1] = 0;
  }
}
int main() {
  long long k;
  scanf("%d %I64d", &n, &k);
  vector<pair<int, int> > A;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    A.push_back({a[i], i});
  }
  sort(A.begin(), A.end());
  siz = 0;
  for (int i = 0; i < n; i++) {
    int j = i;
    while (A[j].first == A[i].first) {
      f[A[j].second] = siz;
      j++;
    }
    i = j - 1;
    siz++;
  }
  for (int i = 0; i < n; i++) {
    int j = i;
    while (A[i].first == A[j].first) {
      F[A[j].second] = siz - 1 - f[A[j].second];
      j++;
    }
    i = j - 1;
  }
  for (int i = 0; i < n; i++) {
    pref[i] = pref[i - 1] + ask(F[i], 0);
    add(F[i] + 1, 0, 1);
  }
  cler();
  for (int i = n - 1; i >= 0; i--) {
    suff[i] = suff[i + 1] + ask(f[i], 0);
    add(f[i] + 1, 0, 1);
  }
  for (int i = 0; i < n; i++) {
  }
  cler();
  add(f[n - 1], 1, 1);
  long long ans = 0;
  for (int i = 0; i < n - 1; i++) {
    add(F[i], 0, 1);
    ans += ask(f[i] - 1, 1);
  }
  int j = n - 2;
  long long res = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (i != n - 1) {
      add(f[i], 1, 1);
      ans += ask(F[i] - 1, 0);
    }
    while ((suff[i] + pref[j] + ans > k || j >= i) && j >= 0) {
      add(F[j], 0, -1);
      ans -= ask(f[j] - 1, 1);
      j--;
    }
    if (j < 0) {
      break;
    }
    res += (j + 1);
  }
  cout << res << endl;
}
