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 maxn = 100000 + 5;
long long t1[400000 + 5], t2[400000 + 5];
long long A[maxn];
long long B[maxn];
long long ans, n;
struct node {
  int id, d, t;
} data[maxn];
int lowbit(int &x) { return x & (-x); }
struct TreeArray {
  int tree[maxn], n;
  inline void add(int a, int b) {
    for (; a < maxn; a += lowbit(a)) tree[a] += b;
  }
  inline int sum(int l, int r) {
    if (l > r) return 0;
    int res = 0;
    l--;
    while (r > l) res += tree[r], r -= lowbit(r);
    while (l > r) res -= tree[l], l -= lowbit(l);
    return res;
  }
} S, T;
inline bool cmp1(node a, node b) { return a.d < b.d; }
inline bool cmp2(node a, node b) { return a.id < b.id; }
int main() {
  int n;
  long long k, ans = 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> data[i].d, data[i].id = i;
  sort(data + 1, data + n + 1, cmp1);
  data[n + 1].t = -1;
  for (int i = 1, j = 1; i <= n; i++, j++) {
    data[i].t = j;
    while (data[i + 1].d == data[i].d) i++, data[i].t = j;
  }
  sort(data + 1, data + n + 1, cmp2);
  long long num = 0;
  for (int i = 1; i <= n; i++)
    num += T.sum(data[i].t + 1, maxn), T.add(data[i].t, 1);
  if (num < k) {
    cout << (long long)n * (n - 1) / 2 << endl;
    return 0;
  }
  int l = 0, r = 1;
  for (l = 1; l <= n; l++) {
    num += S.sum(data[l].t + 1, maxn), num += T.sum(1, data[l].t - 1);
    S.add(data[l].t, 1);
    while (r <= n && (num > k || l == r)) {
      num -= S.sum(data[r].t + 1, maxn), num -= T.sum(1, data[r].t - 1);
      T.add(data[r].t, -1);
      r++;
    }
    if (r > n) break;
    ans += n - r + 1;
  }
  cout << ans << endl;
}
