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 N = 3e5 + 10, MOD = 1e9 + 7, INF = 1e9 + 10, SQ = 550, LOG = 25;
long long n, m, x1, x2, d, z, t, y2, a[N], dp[3][N], mini, ptr, x, y, k, ptr1,
    ptr2, ans;
set<long long> s;
map<long long, long long> mp;
void add(long long x, long long y, long long z) {
  for (x; x < N; x += x & (-x)) dp[z][x] += y;
}
long long sum(long long x, long long z) {
  long long answer = 0;
  for (; x; x -= x & (-x)) answer += dp[z][x];
  return answer;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i], s.insert(a[i]);
  while (s.size()) {
    x = *s.begin();
    mp[x] = ++d;
    s.erase(s.begin());
  }
  for (int i = 1; i <= n; i++) a[i] = mp[a[i]];
  for (int i = n; i >= 2; i--) {
    add(a[i], 1, 2);
    ans += sum(a[i] - 1, 2);
  }
  ptr = 2;
  for (int i = 1; i <= n; i++) {
    add(a[i], 1, 1);
    ans += sum(N - 1, 1) - sum(a[i], 1);
    ans += sum(a[i] - 1, 2);
    while (ptr <= i || (ptr <= n && ans > m)) {
      ans -= sum(N - 1, 1) - sum(a[ptr], 1);
      ans -= sum(a[ptr] - 1, 2);
      add(a[ptr], -1, 2);
      ptr++;
    }
    if (ptr <= n && ans <= m) t += n - ptr + 1;
  }
  cout << t;
}
