#include <bits/stdc++.h>
#define ll long long
using namespace std;


// Các số trong xâu nhỏ hơn 10^5 nên sàng cho nhanh :)
const int Lim = 1e5;
vector<int> pr;
vector<int> lpf;
void Linear_sieve(int n = Lim) {
    pr.assign(1, 2);
    lpf.assign(n + 1, 2);

    for (int x = 3; x <= n; x += 2) {
        if (lpf[x] == 2) pr.push_back(lpf[x] = x);
        for (int i = 1; i < pr.size() && pr[i] <= lpf[x] && pr[i] * x <= n; ++i)
            lpf[pr[i] * x] = pr[i];
    }
}

bool isPrime(int x) {
    if (x < 2) return false;
    return lpf[x] == x;
}

auto solve() {
    int n, x, cnt;
    cin >> n >> x;
    
    while (n--) {
        string ss;
        cin >> ss; 
        ss += "#";
        int num = 0, r = 0;
        
        for (char c: ss) {
            if ('0' <= c && c <= '9') num *= 10, num += c - 48; else r += isPrime(num), num = 0;
        }
        cnt += isPrime(r);
    }
    cout << (cnt >= x ? "YES" : "NO");
}

int main() {
    Linear_sieve();
    
    int t = 1;
    while (t--) solve();
}

const xd = trunc(1e5);
var p:array[0..xd] of boolean;
    i,j,n,x,res,a,c:longint;t,s:string;
procedure sieve;
var i,j:longint;
begin
for i := 2 to trunc(sqrt(xd)) do
 if not p[i] then
   for j := i to xd div i do
           p[i*j] := true;
end;
begin
readln(n);readln(x);sieve;
while n <> 0 do
 begin
   readln(s); s := s + '?'; c:=0;
    for i := 1 to length(s) do
     if s[i] in ['0'..'9'] then t := t + s[i]
     else if length(t)>0 then
       begin
         val(t,a);
         if (not p[a]) and (a>1) then inc(c);
         t := '';
       end;
     if (not p[c]) and (c>1) then inc(res);
     dec(n);
 end;
if res >= x then writeln('YES') else writeln('NO');
end.