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 DIM = 1e5 + 5;
int V[DIM], N;
vector<int> Aux;
long long Inv, K, Ans;
long long A[2][4 * DIM];
void update(int k, int nod, int st, int dr, int p, int x) {
  int Nst = (nod << 1);
  int Ndr = ((nod << 1) | 1);
  if (st == dr) {
    A[k][nod] += x;
  } else {
    int mid = (st + dr) >> 1;
    if (p <= mid)
      update(k, Nst, st, mid + 0, p, x);
    else
      update(k, Ndr, mid + 1, dr, p, x);
    A[k][nod] = A[k][Nst] + A[k][Ndr];
  }
}
long long Sol;
void query(int k, int nod, int st, int dr, int a, int b) {
  if (st > dr || a > b) return;
  int Nst = (nod << 1);
  int Ndr = ((nod << 1) | 1);
  if (a <= st && dr <= b) {
    Sol += A[k][nod];
  } else {
    int mid = (st + dr) >> 1;
    if (a <= mid) query(k, Nst, st, mid + 0, a, b);
    if (b > mid) query(k, Ndr, mid + 1, dr, a, b);
  }
}
int get_norm(int x) {
  int st = 0, dr = Aux.size() - 1;
  while (st <= dr) {
    int mid = (st + dr) >> 1;
    if (x == Aux[mid]) return mid + 1;
    if (Aux[mid] < x)
      st = mid + 1;
    else
      dr = mid - 1;
  }
}
int main() {
  cin >> N >> K;
  for (int i = 1; i <= N; i++) {
    cin >> V[i];
    Aux.push_back(V[i]);
  }
  sort(Aux.begin(), Aux.end());
  Aux.resize(unique(Aux.begin(), Aux.end()) - Aux.begin());
  int M = Aux.size();
  Inv = 0;
  for (int i = N; i >= 2; i--) {
    int x = get_norm(V[i]);
    Sol = 0, query(1, 1, 1, M, 1, x - 1), Inv += Sol;
    update(1, 1, 1, M, x, 1);
  }
  for (int L = 1, R = 2; L < N; L++) {
    if (L == R) {
      int x = get_norm(V[R]);
      Sol = 0, query(1, 1, 1, M, 1, x - 1), Inv -= Sol;
      Sol = 0, query(0, 1, 1, M, x + 1, M), Inv -= Sol;
      update(1, 1, 1, M, x, -1);
      R++;
    }
    int x = get_norm(V[L]);
    Sol = 0, query(1, 1, 1, M, 1, x - 1), Inv += Sol;
    Sol = 0, query(0, 1, 1, M, x + 1, M), Inv += Sol;
    update(0, 1, 1, M, x, 1);
    while (Inv > K && R < N) {
      int x = get_norm(V[R]);
      Sol = 0, query(1, 1, 1, M, 1, x - 1), Inv -= Sol;
      Sol = 0, query(0, 1, 1, M, x + 1, M), Inv -= Sol;
      update(1, 1, 1, M, x, -1);
      R++;
    }
    if (Inv <= K) Ans += (N - R + 1);
  }
  cout << Ans << endl;
  return 0;
}
