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>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
template <typename T>
inline T Abs(T x) {
  return (x >= 0) ? x : -x;
}
template <typename T>
inline T sqr(T x) {
  return x * x;
}
const int INF = (int)1E9;
const long long INF64 = (long long)1E18;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
const int MAXN = 400100;
const int OSN = (1 << 17) - 1;
int n, a[MAXN], ts[MAXN], tp[MAXN];
vector<int> X;
long long k;
int get(int* tree, int v, int l, int r, int ll, int rr) {
  if (l > rr || ll > r || ll > rr) return 0;
  if (ll <= l && r <= rr) return tree[v];
  int mid = (l + r) >> 1;
  return get(tree, 2 * v + 1, l, mid, ll, rr) +
         get(tree, 2 * v + 2, mid + 1, r, ll, rr);
}
void update(int* tree, int v, int value) {
  v += OSN;
  tree[v] += value;
  while (v > 0) {
    v = (v - 1) >> 1;
    tree[v] += value;
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < (int)(n); i++) {
    scanf("%d", &a[i]);
    X.push_back(a[i]);
  }
  sort((X).begin(), (X).end());
  X.erase(unique((X).begin(), (X).end()), X.end());
  for (int i = 0; i < (int)(n); i++)
    a[i] = (int)(lower_bound((X).begin(), (X).end(), a[i]) - X.begin());
  long long cur = 0;
  for (int i = 0; i < (int)(n); i++) {
    cur += get(ts, 0, 0, OSN, a[i] + 1, OSN);
    update(ts, a[i], +1);
  }
  int r = 0;
  long long ans = 0;
  for (int l = 0; l < (int)(n); l++) {
    if (l == r) {
      update(ts, a[l], -1);
      r++;
    } else
      cur +=
          get(tp, 0, 0, OSN, a[l] + 1, OSN) + get(ts, 0, 0, OSN, 0, a[l] - 1);
    update(tp, a[l], +1);
    while (r < n - 1 && cur > k) {
      cur -=
          get(tp, 0, 0, OSN, a[r] + 1, OSN) + get(ts, 0, 0, OSN, 0, a[r] - 1);
      update(ts, a[r], -1);
      r++;
    }
    if (cur <= k) ans += n - r;
  }
  cout << ans << endl;
  return 0;
}
