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 = 1e5 + 7;
const int M = 5;
const int inf = 777777777;
const long long base = 1e18;
const double pi = acos(-1);
const double ep = 1e-9;
int a[N];
long long cnt = 0;
long long ans = 0;
int n, m;
map<int, int> st;
int L[N];
int R[N];
int dp[N];
int get(int x) {
  int ret = 0;
  for (int t = x + 1; t <= m; t += t & -t) ret += L[t];
  return ret;
}
int cnting(int x) {
  int ret = 0;
  for (int t = x + 1; t <= m; t += t & -t) ret += L[t];
  for (int t = x - 1; t > 0; t -= t & -t) ret += R[t];
  return ret;
}
void changeRight(int x, int val) {
  for (int t = x; t <= m; t += t & -t) R[t] += val;
}
void changeLeft(int x, int val) {
  for (int t = x; t > 0; t -= t & -t) L[t] += val;
}
long long k;
int main() {
  cin >> n >> k;
  for (int i = (int)1; i < (int)n + 1; i++) scanf("%d", a + i);
  for (int i = (int)1; i < (int)n + 1; i++) st[a[i]] = 0;
  cnt = 0;
  for (map<int, int>::iterator it = st.begin(); it != st.end(); it++) {
    it->second = ++cnt;
  }
  for (int i = (int)1; i < (int)n + 1; i++) a[i] = st[a[i]];
  m = st.size();
  cnt = ans = 0;
  int j = n + 1;
  for (int i = 1; i <= n; i++) {
    cnt += cnting(a[i]);
    changeLeft(a[i], 1);
    if (i == j) {
      cnt -= cnting(a[j]);
      changeRight(a[j++], -1);
    }
    while (j - 1 > i) {
      int tmp = cnting(a[j - 1]);
      if (cnt + tmp <= k) {
        changeRight(a[--j], 1);
        cnt += tmp;
      } else
        break;
    }
    while (j <= n) {
      int tmp = cnting(a[j]);
      if (cnt > k) {
        cnt -= tmp;
        changeRight(a[j++], -1);
      } else
        break;
    }
    if (j <= n) {
      ans += n - j + 1;
    }
  }
  cout << ans;
}
