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 MAX = 100000 + 10;
int n, a[MAX];
long long K, ans;
struct Li {
  int b, num;
} L[MAX];
int operator<(const Li& a, const Li& b) { return a.b < b.b; }
int cnt, root[MAX];
struct node {
  int lc, rc, sum;
  node() { lc = rc = sum = 0; }
} tree[MAX * 20];
void insert(int old, int u, int l, int r, int x) {
  tree[u] = tree[old];
  if (l == r) {
    ++tree[u].sum;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid) {
    tree[u].lc = ++cnt;
    insert(tree[old].lc, tree[u].lc, l, mid, x);
  } else {
    tree[u].rc = ++cnt;
    insert(tree[old].rc, tree[u].rc, mid + 1, r, x);
  }
  tree[u].sum = tree[tree[u].lc].sum + tree[tree[u].rc].sum;
}
int ask(int u, int l, int r, int a, int b) {
  if (!u) return 0;
  if (r < a || l > b) return 0;
  if (a <= l && r <= b) return tree[u].sum;
  int mid = (l + r) >> 1, ans = 0;
  ans += ask(tree[u].lc, l, mid, a, b);
  ans += ask(tree[u].rc, mid + 1, r, a, b);
  return ans;
}
int ask(int a, int b, int c, int d) {
  return ask(root[b], 1, n, c, d) - ask(root[a - 1], 1, n, c, d);
}
void newroot(int c, int i) {
  root[i] = ++cnt;
  insert(root[i - 1], root[i], 1, n, c);
}
int main() {
  int i, top;
  cin >> n >> K;
  for (i = 1; i <= n; ++i) {
    cin >> a[i];
    L[i].b = a[i];
    L[i].num = i;
  }
  sort(L + 1, L + n + 1);
  for (top = 0, i = 1; i <= n; ++i) {
    if (L[i].b != L[i - 1].b) ++top;
    a[L[i].num] = top;
  }
  for (i = 1; i <= n; ++i) newroot(a[i], i);
  int last = 2;
  long long now = 0;
  for (i = last + 1; i <= n; ++i) now += ask(last, i - 1, a[i] + 1, n);
  for (i = 1; i <= n; ++i) {
    now += ask(last, n, 1, a[i] - 1) + ask(1, i - 1, a[i] + 1, n);
    while (last <= n && (now > K || last == i)) {
      now -= ask(1, i, a[last] + 1, n) + ask(last + 1, n, 1, a[last] - 1);
      ++last;
    }
    if (last == n + 1) break;
    ans += n - last + 1;
  }
  cout << ans << endl;
}
