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 MAX_N = 100000;
static inline int lsb(int x) { return x & (-x); }
struct InversionCounter {
  int aib[2][1 + MAX_N];
  void update(int t, int pos, int val) {
    while (pos <= MAX_N) {
      aib[t][pos] += val;
      pos += lsb(pos);
    }
  }
  int query(int t, int pos) {
    int res = 0;
    while (pos > 0) {
      res += aib[t][pos];
      pos -= lsb(pos);
    }
    return res;
  }
} big_aib;
long long inversions;
void my_insert(int t, int val) {
  inversions =
      inversions + big_aib.query(0, MAX_N - val) + big_aib.query(1, val - 1);
  if (t == 0) val = MAX_N - val + 1;
  big_aib.update(t, val, 1);
}
void my_erase(int t, int val) {
  inversions =
      inversions - big_aib.query(0, MAX_N - val) - big_aib.query(1, val - 1);
  if (t == 0) val = MAX_N - val + 1;
  big_aib.update(t, val, -1);
}
int v[1 + MAX_N];
int* pnt[MAX_N];
bool cmp(int* a, int* b) { return *a < *b; }
void normalize(int n) {
  for (int i = 1; i <= n; ++i) pnt[i - 1] = &v[i];
  std::sort(pnt, pnt + n, cmp);
  int last = *pnt[0], j = 1;
  for (int i = 0; i < n; ++i) {
    if (last == *pnt[i])
      *pnt[i] = j;
    else {
      last = *pnt[i];
      *pnt[i] = ++j;
    }
  }
}
int main() {
  int n;
  long long k;
  scanf("%d%lld", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &v[i]);
  normalize(n);
  for (int i = 1; i <= n - 1; ++i) my_insert(0, v[i]);
  int last = n - 1;
  long long res = 0LL;
  for (int i = n; i >= 2; --i) {
    my_insert(1, v[i]);
    if (last == i) {
      my_erase(0, v[last]);
      last--;
    }
    while (last > 0 && inversions > k) {
      my_erase(0, v[last]);
      --last;
    }
    res = res + last;
  }
  printf("%lld", res);
  return 0;
}
