#ifdef __GNUC__
#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#endif

#define _USE_MATH_DEFINES

#include <iostream>
#include <iomanip>

#include <sstream>
#include <algorithm>
#include <cmath>

#include <string>
#include <cstring>
#include <vector>
#include <valarray>
#include <array>

#include <queue>
#include <complex>
#include <set>
#include <map>
#include <stack>
#include <list>

#include<cassert>//assert();
#include <fstream>
/////////
#define REP(i, x, n) for(int i = x; i < n; i++)
#define rep(i,n) REP(i,0,n)
#define P(p) cout<<(p)<<endl;
 
#define PII pair<int,int>
#define MAIN_INIT std::cin.tie(0); \
    std::ios::sync_with_stdio(false);\
	std::cout << std::fixed;//小数を10進数表示
/////////
#ifdef getchar_unlocked
#define mygc(c) (c)=getchar_unlocked()
#else
#define mygc(c) (c)=getchar()
#endif
#ifdef putchar_unlocked
#define mypc(c) putchar_unlocked(c)
#else
#define mypc(c) putchar(c)
#endif
/////////
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
/////////
using namespace::std;
/////////
#ifdef _DEBUG
#define DEBUG_BOOL(b) assert(b)
#else
#define DEBUG_BOOL(b) 
#endif
/////数値読み込み
#define ENABLE_READER_ON(T)	\
inline void reader(T &x){int k;x = 0;bool flag = true;\
while(true){mygc(k);\
if( k == '-'){flag = false;break;}if('0' <= k && k <= '9'){x = k - '0';break;}\
}\
if( flag ){while(true){mygc(k);if( k<'0' || '9'<k)break;x = x * 10 + (k - '0');}}\
else{while(true){mygc(k);if( k<'0' || '9'<k)break;x = x * 10 - (k-'0');}}\
}
//整数
ENABLE_READER_ON(int)
ENABLE_READER_ON(long)
ENABLE_READER_ON(long long)
ENABLE_READER_ON(unsigned int)
ENABLE_READER_ON(unsigned long)
ENABLE_READER_ON(unsigned long long)

//////
//文字読み込み
inline int reader(char& c){int i;
for(;;){mygc(i);if(i!=' '&&i!='\n'&&i!='\r'&&i!='\t'&&i!=EOF) break;}
c = i;
return i;}

inline int reader(char c[]){int i,s=0;
for(;;){mygc(i);if(i!=' '&&i!='\n'&&i!='\r'&&i!='\t'&&i!=EOF) break;}
c[s++]=i;
for(;;){mygc(i);if(i==' '||i=='\n'||i=='\r'||i=='\t'||i==EOF) break;c[s++]=i;}
c[s]='\0';return s;
}

inline int reader(string& c,int size=100){int i;c.reserve(size);
for(;;){mygc(i);if(i != ' '&&i != '\n'&&i != '\r'&&i != '\t'&&i != EOF)break;}
c.push_back(i);
for(;;){mygc(i);if(i == ' ' || i == '\n' || i == '\r' || i == '\t' || i == EOF)break;c.push_back(i);}
return c.size();}

/////////
//数値出力
#define ENABLE_WRITER_ON(T)	\
inline void writer(T x){char f[20];int s = 0;\
if (x<0){mypc('-');while(x){f[s++] = ~(x%10)+1,x /= 10;}}\
else{while(x){f[s++] = (x % 10), x /= 10;}}\
if (!s)f[s++] = 0;while (s--)mypc(f[s] + '0');}

ENABLE_WRITER_ON(int)
ENABLE_WRITER_ON(long)
ENABLE_WRITER_ON(long long)
ENABLE_WRITER_ON(unsigned int)
ENABLE_WRITER_ON(unsigned long)
ENABLE_WRITER_ON(unsigned long long)
/////////
inline void writer(const char c){mypc(c);}
inline void writer(const char c[]){for (int i = 0; c[i] != '\0'; i++)mypc(c[i]); }
inline void writer(const string str){writer( str.c_str() );}
///////////////////////////////////////////////////////////
// 最大公約数
template<class T>
inline T gcd(T a, T b){return b == 0 ? a : gcd(b, a % b);}
// 最小公倍数
template<class T>
inline T lcm(T a, T b){return a * b / gcd(a, b);}
////////////////////////////////

int primeMax;
vector<int> prime;
vector<bool> flag;
void makePrime(int N){
	flag = vector<bool>(N+1,true);
	prime.push_back(2);
	for(int i=4;i<=N;i+=2){
		flag[i] = false;
	}
	for(int i=3;i<=N;i+=2){
		if( flag[i] == true ){
			for(int j=i+i;j<=N;j+=i){
				flag[j] = false;
			}
			prime.push_back( i );
		}
	}
	primeMax = prime.size();
}

inline void solve(){
	int N;
	reader(N);//20000
	makePrime(N);

	vector<int> dp(N+1,0);//dp[i]:数iを作る、最大和回数。
	dp[0] = 1;//0を和1回で生成する。最後に1引く
	
	//prime[primeIndex]で数を作る。	
	for(int primeIndex = 0;primeIndex < primeMax;++primeIndex ){
		//prime[primeIndex]を足すので、dpの後ろから作る。
		for(int number= N;number >= 0;--number){
			if( dp[number] == 0 ){continue;}//作られていない
			if( prime[primeIndex] + number > N ){continue;}//調べる値を超えている。
			int terNum;
			//作る数
			terNum = number + prime[primeIndex];
			//和の回数が大きい値を選択する。
			dp[terNum] = max(dp[terNum], dp[number]+1);
		}
	}
	/*
	//writer(primeMax);writer("\n");
	N=20000で2262個の素数。
	2262*20000=45240000=4.524*10**7
	*/
	
	writer( dp[N]-1 );//0の生成分の和回数1を引く
}

int main(void){
    MAIN_INIT;
    cout << setprecision(16);//小数をいっぱい表示する。16?

	solve();

	return 0;
}