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;
char buf[1 << 21], *p1 = buf, *p2 = buf;
template <class T>
void read(T &x) {
  x = 0;
  int c = getchar();
  int flag = 0;
  while (c < '0' || c > '9') flag |= (c == '-'), c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (flag) x = -x;
}
template <class T>
T _max(T a, T b) {
  return b < a ? a : b;
}
template <class T>
T _min(T a, T b) {
  return a < b ? a : b;
}
template <class T>
bool checkmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
bool checkmin(T &a, T b) {
  return b < a ? a = b, 1 : 0;
}
const int N = 100005;
int n;
long long k;
int a[N], arr[N], top = 0;
long long inv[N];
struct T1 {
  int C[N];
  void update(int x, int num) {
    while (x) {
      C[x] += num;
      x -= (x & (-x));
    }
  }
  int query(int x) {
    int ans = 0;
    while (x <= top) {
      ans += C[x];
      x += (x & (-x));
    }
    return ans;
  }
} S;
struct T2 {
  int C[N];
  void update(int x, int num) {
    while (x <= top) {
      C[x] += num;
      x += (x & (-x));
    }
  }
  int query(int x) {
    int ans = 0;
    while (x) {
      ans += C[x];
      x -= (x & (-x));
    }
    return ans;
  }
} T;
void init() {
  read(n);
  read(k);
  for (int i = 1; i <= n; ++i) read(a[i]), arr[++top] = a[i];
  sort(arr + 1, arr + top + 1);
  top = unique(arr + 1, arr + top + 1) - arr - 1;
  for (int i = 1; i <= n; ++i)
    a[i] = lower_bound(arr + 1, arr + top + 1, a[i]) - arr;
}
void solve() {
  long long ans = 0, tot = 0;
  for (int i = n; i >= 1; --i) {
    tot += T.query(a[i] - 1);
    T.update(a[i], 1);
  }
  for (int l = 1, r = 1; l < n; ++l) {
    if (l == r) {
      tot -= S.query(a[r] + 1);
      T.update(a[r], -1);
      tot -= T.query(a[r] - 1);
      ++r;
    }
    tot += S.query(a[l] + 1);
    tot += T.query(a[l] - 1);
    S.update(a[l], 1);
    while (tot > k && r <= n) {
      tot -= S.query(a[r] + 1);
      T.update(a[r], -1);
      tot -= T.query(a[r] - 1);
      ++r;
    }
    ans += n - r + 1;
  }
  printf("%lld\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}
