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 len = 1e5 + 9;
int n, cou = 1;
;
long long k;
int a[len];
set<int> s;
map<int, int> m;
int last = (1 << 17);
long long finalres = 0;
struct segmentTree {
  struct node {
    long long cou = 0;
  };
  node tre[(1 << 18)];
  void update(int ind, int type, int now = 1, int be = 1, int ed = last) {
    if (be == ed) {
      tre[now].cou += 1 * type;
      return;
    }
    int mid = (be + ed) / 2;
    if (ind <= mid) {
      update(ind, type, now * 2, be, mid);
    } else {
      update(ind, type, now * 2 + 1, mid + 1, ed);
    }
    tre[now].cou = tre[now * 2].cou + tre[now * 2 + 1].cou;
  }
  long long query(int l, int r, int now = 1, int be = 1, int ed = last) {
    if (l > r) return 0;
    if (be == ed) {
      return tre[now].cou;
    }
    if (be >= l && ed <= r) {
      return tre[now].cou;
    }
    int mid = (be + ed) / 2;
    long long res = 0;
    if (l <= mid) {
      res += query(l, r, now * 2, be, mid);
    }
    if (r > mid) {
      res += query(l, r, now * 2 + 1, mid + 1, ed);
    }
    return res;
  }
};
int main() {
  scanf("%d%I64d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    s.insert(a[i]);
  }
  for (set<int>::iterator it = s.begin(); it != s.end(); it++) {
    m[*it] = cou++;
  }
  segmentTree le;
  segmentTree ri;
  long long cur = 0;
  int inde;
  le.update(m[a[0]], 1);
  for (int i = 1; i < n; i++) {
    inde = m[a[i]];
    cur += ri.query(inde + 1, last);
    if (a[i] < a[0]) cur++;
    ri.update(inde, 1);
  }
  if (cur <= k) {
    finalres = n;
    finalres *= n - 1;
    finalres /= 2;
    printf("%I64d", finalres);
    return 0;
  }
  int be = 0, ed = 1;
  while (ed < n && be < ed) {
    if (cur > k) {
      inde = m[a[ed]];
      cur -= ri.query(1, inde - 1);
      cur -= le.query(inde + 1, last);
      ri.update(inde, -1);
      ed++;
    } else {
      finalres += ed - be - 1;
      be++;
      inde = m[a[be]];
      le.update(inde, 1);
      cur += le.query(inde + 1, last);
      cur += ri.query(1, inde - 1);
    }
  }
  long long test = (ed - be);
  finalres += (test) * (test - 1) / 2;
  long long fin = n;
  fin *= n - 1;
  fin /= 2;
  printf("%I64d", fin - finalres);
}
