#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = (int)1e6 + 100;
int n, k, a[N], pre[N], t[N], ans = 0, l[N], r[N];
int get(int m, int n) { return (int)lround(100.0 * m / n); }
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    int id = 1;
    for (int j = 1; j <= k; ++j) {
      if (t[j] < t[id]) id = j;
    }
    l[i] = t[id];
    r[i] = t[id] + a[i];
    t[id] += a[i];
    pre[t[id]]++;
  }
  for (int i = 1; i < N; ++i) {
    pre[i] += pre[i - 1];
  }
  for (int i = 1; i <= n; ++i) {
    bool ok = false;
    for (int j = l[i]; j < r[i]; ++j) {
      if (get(pre[j], n) == j - l[i] + 1) ok = true;
    }
    ans += ok;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

Vasya likes taking part in Codeforces contests. When a round is over, Vasya follows all submissions in the system testing tab.

There are n solutions, the i-th of them should be tested on a_i tests, testing one solution on one test takes 1 second. The solutions are judged in the order from 1 to n. There are k testing processes which test solutions simultaneously. Each of them can test at most one solution at a time.

At any time moment t when some testing process is not judging any solution, it takes the first solution from the queue and tests it on each test in increasing order of the test ids. Let this solution have id i, then it is being tested on the first test from time moment t till time moment t + 1, then on the second test till time moment t + 2 and so on. This solution is fully tested at time moment t + a_i, and after that the testing process immediately starts testing another solution.

Consider some time moment, let there be exactly m fully tested solutions by this moment. There is a caption "System testing: d%" on the page with solutions, where d is calculated as

$$$d = round\left(100⋅m/n\right),$$$

where round(x) = ⌊{x + 0.5}⌋ is a function which maps every real to the nearest integer.

Vasya calls a submission interesting if there is a time moment (possibly, non-integer) when the solution is being tested on some test q, and the caption says "System testing: q%". Find the number of interesting solutions.

Please note that in case when multiple processes attempt to take the first submission from the queue at the same moment (for instance, at the initial moment), the order they take the solutions does not matter.

Input

The first line contains two positive integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ 100) standing for the number of submissions and the number of testing processes respectively.

The second line contains n positive integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 150), where a_i is equal to the number of tests the i-th submission is to be run on.

Output

Output the only integer — the number of interesting submissions.

Examples

Input


2 2
49 100


Output


1


Input


4 2
32 100 33 1


Output


2


Input


14 5
48 19 6 9 50 20 3 42 38 43 36 21 44 6


Output


5

Note

Consider the first example. At time moment 0 both solutions start testing. At time moment 49 the first solution is fully tested, so at time moment 49.5 the second solution is being tested on the test 50, and the caption says "System testing: 50%" (because there is one fully tested solution out of two). So, the second solution is interesting.

Consider the second example. At time moment 0 the first and the second solutions start testing. At time moment 32 the first solution is fully tested, the third solution starts testing, the caption says "System testing: 25%". At time moment 32 + 24.5 = 56.5 the third solutions is being tested on test 25, the caption is still the same, thus this solution is interesting. After that the third solution is fully tested at time moment 32 + 33 = 65, the fourth solution is fully tested at time moment 65 + 1 = 66. The captions becomes "System testing: 75%", and at time moment 74.5 the second solution is being tested on test 75. So, this solution is also interesting. Overall, there are two interesting solutions.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = (int)1e6 + 100;
int n, k, a[N], pre[N], t[N], ans = 0, l[N], r[N];
int get(int m, int n) { return (int)lround(100.0 * m / n); }
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    int id = 1;
    for (int j = 1; j <= k; ++j) {
      if (t[j] < t[id]) id = j;
    }
    l[i] = t[id];
    r[i] = t[id] + a[i];
    t[id] += a[i];
    pre[t[id]]++;
  }
  for (int i = 1; i < N; ++i) {
    pre[i] += pre[i - 1];
  }
  for (int i = 1; i <= n; ++i) {
    bool ok = false;
    for (int j = l[i]; j < r[i]; ++j) {
      if (get(pre[j], n) == j - l[i] + 1) ok = true;
    }
    ans += ok;
  }
  cout << ans << endl;
  return 0;
}

```