stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;



typedef long long ll;



#define faster ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);



#define endl "\n"



int main()

{

	faster;

	

	int n;

	cin >> n;

	bool contains = false;

	while(n > 0){

		if(n%10 == 7) contains = true;

		n /= 10;

	}



	cout << (contains ? "Yes" : "No") << endl;

	

	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

	char str[4];

	scanf("%s", str);

	if (str[0] == '7' || str[1] == '7' || str[2] == '7') {

		printf("Yes\n");

	}

	else {

		printf("No\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int main()

{

    string num;



    cin >> num;



    for(int i=0; i<num.size(); i++)

    {

        if(num[i] == '7')

        {

            cout << "Yes";

            return 0;

        }

    }

    

    cout << "No";





}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

	char str[4];

	scanf("%s", str);

	if (str[0] == '7' || str[1] == '7' || str[2] == '7') {

		printf("Yes\n");

	}

	else {

		printf("No\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using  namespace  std;



int main() {



    int n;

    cin>>n;



    bool flag = false;

    while(n){

        int rem = n%10;

        if(rem==7){

            flag = true;

            break;

        }

        n = n/10;

    }



    if(flag){

        cout<<"Yes\n";

    }

    else{

        cout<<"No\n";

    }

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

	char str[4];

	scanf("%s", str);

	if (str[0] == '7' || str[1] == '7' || str[2] == '7') {

		printf("Yes\n");

	}

	else {

		printf("No\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>



using namespace std;



int main()

{

    string line;

    cin >> line;



    for(int i=0; i<line.size(); i++)

    {

        if(line[i]=='7')

        {

            cout<<"Yes\n";

            return 0;

        }

    }



    cout<<"No\n";



    return 0;



}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

	char str[4];

	scanf("%s", str);

	if (str[0] == '7' || str[1] == '7' || str[2] == '7') {

		printf("Yes\n");

	}

	else {

		printf("No\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define io ios_base::sync_with_stdio(false); cin.tie(NULL);



#define N 10000000

#define nl "\n"

#define pb push_back

#define MOD 1000000007

#define pi acos(-1)

#define f first

#define s second



typedef long long ll;

typedef double dd;

//typedef vector<int> vi;

//typedef vector<ll> vll;



using namespace std;



bool seven(int x)

{

    int mod;

    while(x!=0)

    {

        mod = x%10;

        if(mod == 7)

        {

            return true;

            break;

        }

        x = x/ 10;

    }

    return false;

}



int main()

{

    io;

    //freopen("in.txt" , "r" , stdin);

    //freopen("out.txt" , "w" , stdout);



    ///Let the showdown begin



    int t;

    cin >> t;



    if(seven(t))

        cout << "Yes" << nl;

    else

        cout << "No" << nl;



    /// Atlast DONE 

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

	char str[4];

	scanf("%s", str);

	if (str[0] == '7' || str[1] == '7' || str[2] == '7') {

		printf("Yes\n");

	}

	else {

		printf("No\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

string n;

int main()

{

    int i,c=0;

    cin>>n;

    for(i=0;i<n.length();i++)

    {

        if(n[i]=='7')

        {

            cout<<"Yes\n";

            c=1;

            break;

        }

    }

    if(c==0)

    {

        cout<<"No\n";

    }



return 0;

}




// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

	char str[4];

	scanf("%s", str);

	if (str[0] == '7' || str[1] == '7' || str[2] == '7') {

		printf("Yes\n");

	}

	else {

		printf("No\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
// writer: w33z8kqrqk8zzzx33

#include <bits/stdc++.h>

using namespace std;



// begin fast read template by CYJian (source: https://www.luogu.com.cn/paste/i11c3ppx)



namespace io {

	const int __SIZE = (1 << 21) + 1;

	char ibuf[__SIZE], *iS, *iT, obuf[__SIZE], *oS = obuf, *oT = oS + __SIZE - 1, __c, qu[55]; int __f, qr, _eof;

	#define Gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, __SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)

	inline void flush () { fwrite (obuf, 1, oS - obuf, stdout), oS = obuf; }

	inline void gc (char &x) { x = Gc(); }

	inline void pc (char x) { *oS ++ = x; if (oS == oT) flush (); }

	inline void pstr (const char *s) { int __len = strlen(s); for (__f = 0; __f < __len; ++__f) pc (s[__f]); }

	inline void gstr (char *s) { for(__c = Gc(); __c < 32 || __c > 126 || __c == ' ';)  __c = Gc();

		for(; __c > 31 && __c < 127 && __c != ' '; ++s, __c = Gc()) *s = __c; *s = 0; }

	template <class I> inline bool gi (I &x) { _eof = 0;

		for (__f = 1, __c = Gc(); (__c < '0' || __c > '9') && !_eof; __c = Gc()) { if (__c == '-') __f = -1; _eof |= __c == EOF; }

		for (x = 0; __c <= '9' && __c >= '0' && !_eof; __c = Gc()) x = x * 10 + (__c & 15), _eof |= __c == EOF; x *= __f; return !_eof; }

	template <class I> inline void print (I x) { if (!x) pc ('0'); if (x < 0) pc ('-'), x = -x;

		while (x) qu[++ qr] = x % 10 + '0',  x /= 10; while (qr) pc (qu[qr --]); }

	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;

} using io::pc; using io::gc; using io::pstr; using io::gstr; using io::gi; using io::print;



// end fast read template by CYJian



#define iter(i, a, b) for(int i=(a); i<(b); i++)

#define rep(i, a) iter(i, 0, a)

#define rep1(i, a) iter(i, 1, (a)+1)

#define log(a) cerr<<"\033[32m[DEBUG] "<<#a<<'='<<(a)<<" @ line "<<__LINE__<<"\033[0m"<<endl

#define all(a) a.begin(), a.end()

#define fi first

#define se second

#define pb push_back

#define mp make_pair

 

using ll=long long;

using pii=pair<int, int>;

//#define int ll

const int MOD = 1000000007;



signed main() {

    ios_base::sync_with_stdio(false); cin.tie(0);

    char a, b, c;

	gc(a), gc(b), gc(c);

	if(a == '7' || b == '7' || c == '7') pstr("Yes");

	else pstr("No");

}


// Optimized solution to the same problem (faster):
#include<cstdio>

using namespace std;

int main(){

	char c;

	if(getchar()=='7'){printf("Yes\n");return 0;}

	if(getchar()=='7'){printf("Yes\n");return 0;}

	if(getchar()=='7'){printf("Yes\n");return 0;}

	printf("No\n");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main(){

	char a,b,c;

	cin>>a>>b>>c;

	if(a=='7'||b=='7'||c=='7')cout<<"Yes"<<endl;

	else cout<<"No"<<endl;

	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

using namespace std;

int main(){

	char c;

	if(getchar()=='7'){printf("Yes\n");return 0;}

	if(getchar()=='7'){printf("Yes\n");return 0;}

	if(getchar()=='7'){printf("Yes\n");return 0;}

	printf("No\n");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;

int main(){

    int a;

    cin>>a;

    for(int i=1;i<=3;i++){

    if(a%10==7){

    cout<<"Yes";return 0;

    }

  else a=a/10;

  }

  cout<<"No";//如果没有输出no

}

// Optimized solution to the same problem (faster):
#include<cstdio>

using namespace std;

int main(){

	char c;

	if(getchar()=='7'){printf("Yes\n");return 0;}

	if(getchar()=='7'){printf("Yes\n");return 0;}

	if(getchar()=='7'){printf("Yes\n");return 0;}

	printf("No\n");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;

int main(){

int a;

cin>>a;

  for(int i=1;i<=3;i++)

{

  if(a%10==7){cout<<"Yes";return 0;}

  else a=a/10;}

  cout<<"No";

}

// Optimized solution to the same problem (faster):
#include<cstdio>

using namespace std;

int main(){

	char c;

	if(getchar()=='7'){printf("Yes\n");return 0;}

	if(getchar()=='7'){printf("Yes\n");return 0;}

	if(getchar()=='7'){printf("Yes\n");return 0;}

	printf("No\n");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

string s;

int main()

{

    cin>>s;

    for(int i=0;i<3;i++)

        if(s[i]=='7')

        {

            printf("Yes\n");

            return 0;

        }

    printf("No\n");

    return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

using namespace std;

int main(){

	char c;

	if(getchar()=='7'){printf("Yes\n");return 0;}

	if(getchar()=='7'){printf("Yes\n");return 0;}

	if(getchar()=='7'){printf("Yes\n");return 0;}

	printf("No\n");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;

int main()

{

    char a[5];

    cin>>a;

    for(int i=0;i<3;i++)

    {

        if(a[i]=='7')

        {

            cout<<"Yes"<<endl;

            return 0;

        }

    }

    cout<<"No"<<endl;

    return 0;

}

// Optimized solution to the same problem (faster):
//#include<bits/stdc++.h>

#include<stdio.h>

#include<string.h>

using namespace std;

#define ll long long

const ll maxn = 1e6 + 10;

//const double pi = acos(-1.0);

 

char s[10]; 

int main(void)

{

	scanf("%s",s);

	int len = strlen(s);

	int tp=0;

	for(int i=0;i<len;i++)

	{

		tp = s[i] - '0';

		if(tp==7)break;

	}

	if(tp==7){

		printf("Yes\n");

	//	printf("%s contains 7 as its last digit.\n",s);	

	}

	else 

	{

		printf("No\n");

	//	printf("%s doesdoes not contain the digit 7.\n",s);

	}

	return 0;

} 

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<algorithm>

#include<cmath>

#include<cstdlib>

#include<cstdio>

#include<cstring>

#include<iomanip>

#include<string>

#include<set>

#include<map>

#include<vector>

#include<queue>

#include<stack>

#include<utility>

#define PI acos(-1.0)

using namespace std;

int main(){

	char a[3];

	for(int i =0;i < 3;i++) cin >> a[i];

	int flag = 0;

	for(int i = 0;i < 3;i++){

		if(a[i] == '7'){

			flag = 1;

			break;

		} 

	}

	if(flag) cout << "Yes" << endl;

	else cout << "No" << endl;

	return 0;

}

// Optimized solution to the same problem (faster):
//#include<bits/stdc++.h>

#include<stdio.h>

#include<string.h>

using namespace std;

#define ll long long

const ll maxn = 1e6 + 10;

//const double pi = acos(-1.0);

 

char s[10]; 

int main(void)

{

	scanf("%s",s);

	int len = strlen(s);

	int tp=0;

	for(int i=0;i<len;i++)

	{

		tp = s[i] - '0';

		if(tp==7)break;

	}

	if(tp==7){

		printf("Yes\n");

	//	printf("%s contains 7 as its last digit.\n",s);	

	}

	else 

	{

		printf("No\n");

	//	printf("%s doesdoes not contain the digit 7.\n",s);

	}

	return 0;

} 

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

int main()

{

    string s;  cin>>s;

    if(s.find('7')!=-1)

        cout<<"Yes";

    else

        cout<<"No";

    return 0;

}


// Optimized solution to the same problem (faster):
//#include<bits/stdc++.h>

#include<stdio.h>

#include<string.h>

using namespace std;

#define ll long long

const ll maxn = 1e6 + 10;

//const double pi = acos(-1.0);

 

char s[10]; 

int main(void)

{

	scanf("%s",s);

	int len = strlen(s);

	int tp=0;

	for(int i=0;i<len;i++)

	{

		tp = s[i] - '0';

		if(tp==7)break;

	}

	if(tp==7){

		printf("Yes\n");

	//	printf("%s contains 7 as its last digit.\n",s);	

	}

	else 

	{

		printf("No\n");

	//	printf("%s doesdoes not contain the digit 7.\n",s);

	}

	return 0;

} 

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

string s;

int main(){

	ios::sync_with_stdio(0);

	cin.tie(0);

	cin>>s;

	if(s[0]=='7'||s[1]=='7'||s[2]=='7') return cout<<"Yes",0;

	cout<<"No";

	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;



char s[5];

int n,flag;



int main()

{

	scanf("%s",s);

	n=strlen(s);

	for(int i=0;i<n;i++)

		{

			if(s[i]=='7')

				{

					printf("Yes");

					flag++;

					break;

				}

			

		}

	if(flag==0)printf("No");

	return 0;



	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



int main(){

    int n; cin >> n;

    while(n > 0){

        if(n%10 == 7){

            cout << "Yes\n";

            return 0;

        }

        n /= 10;

    }

    cout << "No\n";

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define FOR(I,A,B)          for(int I=A; I<B; I++)

#define LL                  long long

#define PII                 pair<int,int>

#define MOD                 1000000007

#define MAX                 1000005

#define READ(A)             freopen(A,"r",stdin)



int main() {

    char S[4];

    bool B=false;

    scanf("%s",S);

    FOR(I,0,3){

        if(S[I]=='7'){

            B=true;

        }

    }

    if(B)printf("Yes\n");

    else printf("No\n");

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main(){

    string n;

    int i;

    cin>>n;

    bool yes=false;

    for(i=0;i<3;i++){

        if(n[i]=='7'){

            yes=true;

            cout<<"Yes"<<endl;

            break;

        }



    }

    if(yes==false){

        cout<<"No"<<endl;

    }





    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define mp make_pair

#define ff first

#define ss second

#define pb push_back

#define w(x) scanf("%d",&x);while(x--)

#define rep(i,a,b) for(i=a;i<=b;i++)

#define ios ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define pi 2*acos(0.0)

using namespace std;

typedef long long ll;

typedef pair<ll,ll> pll;

typedef vector<int> vi;



long long int mod=1e9+9;

const int N=100000;

int t=3,i,j,cs=0,n,a,b;

string s;



int main()

{

    cin>>s;

   for(auto it:s)

    if(it=='7'){

        cout<<"Yes"<<endl;

        return 0;

    }

        cout<<"No"<<endl;

}








// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



#define MorsheD ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)

typedef long long ll;



bool mew(int n) {

    while (n) {

        if (n%10 == 7) {

            return true;

        }

        n /= 10;

    }

    return false;

}



int main(){

    // ===================

    MorsheD;

    // freopen("in.txt", "r", stdin);

    // freopen("out.txt", "w", stdout);

    // ===================



    int n;

    cin >> n;



    if (mew(n)) {

        cout << "Yes" << "\n";

    } else {

        cout << "No" << "\n";

    }



    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;



int main() {





    char a[3];

    for(int i=0; i<3; i++){

        cin>>a[i];

        if(a[i]=='7') {cout<<"Yes"<<endl; return 0;}

    }

    cout<<"No"<<endl;

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<string>

using namespace std;



int main()

{

    bool con = false;

    string n;

    cin >> n;



    for(int i = 0; i < n.size(); i++)

    {

        if(n[i] == '7')

            con = true;

    }



    if(con)

        cout << "Yes";

    else

        cout << "No";



    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;



int main()

{

    int n,re,flag;

    while(cin>>n)

    {

        flag=0;

        while(n!=0)

        {

            re=n%10;

            if(re==7)

            {

                flag=1;

                break;

            }

            n=n/10;

        }

        if(flag==1)

            cout<<"Yes";

        else

            cout<<"No";

    }

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>



using namespace std;



typedef long long ll;



int main()

{

    ll a;



    cin>> a;



    ll b = a;



        ll c3 = a %10;

        a /= 10;



        ll c2 = a % 10;

        a/=10;



        ll c1 = a %10;



    if(c1 == 7 || c2 == 7 || c3 == 7) cout<< "Yes\n";



    else cout<< "No\n";



	return 0;

}




// Optimized solution to the same problem (faster):
#include <stdio.h>



int main () {

    int n;

    scanf("%d", &n);



    int first = n / 100;

    int second = (n / 10) % 10;

    int third = n % 10;



    while ( n > 0) {

        if ( first == 7) {

            puts("Yes");

            break;

        }

        else if ( second == 7) {

            puts("Yes");

            break;

        }

        else if ( third == 7) {

            puts("Yes");

            break;

        }

        else {

            puts("No");

            break;

        }

    } 

    return 0;

}

// Fastest known solution for this problem: