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 long long maxn = 1e5 + 5;
long long segsuff[4 * maxn], segpref[4 * maxn], a[maxn];
map<long long, long long> M;
void addsuff(long long id, long long L, long long R, long long idx,
             long long val) {
  if (idx < L || idx >= R) return;
  if (R == L + 1) {
    segsuff[id] += val;
    return;
  }
  long long mid = (L + R) / 2;
  addsuff(2 * id + 0, L, mid, idx, val);
  addsuff(2 * id + 1, mid, R, idx, val);
  segsuff[id] = segsuff[2 * id + 0] + segsuff[2 * id + 1];
}
long long getsuff(long long id, long long L, long long R, long long l,
                  long long r) {
  if (R <= l || r <= L) return 0;
  if (L >= l && R <= r) return segsuff[id];
  long long mid = (L + R) / 2;
  return getsuff(2 * id + 0, L, mid, l, r) + getsuff(2 * id + 1, mid, R, l, r);
}
void addpref(long long id, long long L, long long R, long long idx,
             long long val) {
  if (idx < L || idx >= R) return;
  if (R == L + 1) {
    segpref[id] += val;
    return;
  }
  long long mid = (L + R) / 2;
  addpref(2 * id + 0, L, mid, idx, val);
  addpref(2 * id + 1, mid, R, idx, val);
  segpref[id] = segpref[2 * id + 0] + segpref[2 * id + 1];
}
long long getpref(long long id, long long L, long long R, long long l,
                  long long r) {
  if (R <= l || r <= L) return 0;
  if (L >= l && R <= r) return segpref[id];
  long long mid = (L + R) / 2;
  return getpref(2 * id + 0, L, mid, l, r) + getpref(2 * id + 1, mid, R, l, r);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> compress;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    compress.push_back(a[i]);
  }
  sort(compress.begin(), compress.end());
  compress.resize(unique(compress.begin(), compress.end()) - compress.begin());
  for (long long i = 0; i < compress.size(); i++) {
    M[compress[i]] = i;
  }
  for (long long i = 0; i < n; i++) {
    a[i] = M[a[i]];
  }
  long long ans = 0, inv = 0;
  long long ptr2 = n;
  while (inv <= k && ptr2) {
    if (getsuff(1, 0, n, 0, a[ptr2 - 1]) + inv > k) break;
    ptr2--;
    inv += getsuff(1, 0, n, 0, a[ptr2]);
    addsuff(1, 0, n, a[ptr2], +1);
  }
  if (!ptr2) {
    cout << n * (n - 1ll) / 2ll;
    return 0;
  }
  for (long long ptr1 = 0; ptr1 < n - 1; ptr1++) {
    inv += getpref(1, 0, n, a[ptr1] + 1, n) + getsuff(1, 0, n, 0, a[ptr1]);
    addpref(1, 0, n, a[ptr1], +1);
    while (inv > k && ptr2 < n) {
      inv -= getsuff(1, 0, n, 0, a[ptr2]) + getpref(1, 0, n, a[ptr2] + 1, n);
      addsuff(1, 0, n, a[ptr2], -1);
      ptr2++;
    }
    ans += n - max(ptr2, ptr1 + 1);
  }
  cout << ans;
}
