/**
 * Deon Nicholas' solution to Project Panoptes.
 *
 * It is effectively brute force. Try all possible period lengths L (>= p),
 * and for each period length L, check each possible starting position
 * in [0,L).
 *
 * Stop once we find one. The total running time is O(N^2).
 */ 

#include <iostream>
#include <ctime>
#include <fstream>
#include <cmath>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <complex>
#include <utility>
#include <cctype>
#include <list>
#include <bitset>

using namespace std;

#define FORALL(i,a,b) for(int i=(a);i<=(b);++i)
#define FOR(i,n) for(int i=0;i<(n);++i)
#define FORB(i,a,b) for(int i=(a);i>=(b);--i)

typedef long long ll;
typedef long double ld;
typedef complex<ld> vec;

typedef pair<ll,int> plli;
typedef pair<int,int> pii;
typedef map<int,int> mii;

#define pb push_back
#define mp make_pair

#define MAXN 1005

double A[MAXN];

bool works(int len, int N, double goal) {
  FOR(i,len) {
    bool good = true;
    for(int j = i; j < N; j+=len) {
      if (A[j] >= goal) {
        good = false;
        break;
      }
    }
    if (good) return true;
  }
  return false;
}

int main() {
  int N,P;
  scanf("%d%d",&N,&P);

  assert(2 <= N && N <= 1000);
  assert(1 <= P && P <= N-1);

  double sum = 0;
  FOR(i,N) scanf("%lf",&A[i]), sum += A[i];
  sum /= N;

  double goal = sum * 8 / 10;

  FORALL(len,P,N) if (works(len,N,goal)) {
    printf("%d\n",len);
    return 0;
  }

  printf("-1\n");
}