stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
////////////////////////// Author

////////////////////////// Nasim Hossain Rabbi

////////////////////////// JU-CSE28

////////////////////////// CF - imnasim3.1415

////////////////////////// UVA - imnasim3.1415

////////////////////////// Mail - imnasim3.1415@gmail.com

#include<bits/stdc++.h>

using namespace std;

#define E end()

#define B begin()

#define sz size()

#define EM empty()

#define clr clear()

#define sqr(a) (a*a)

#define mk make_pair

#define pb push_back

string en="\n";

string t="hello";

#define rep(i,n) for(i=0;i<n;i++)

#define Rep(i,n) for(i=1;i<=n;i++)

#define per(i,n) for(i=n-1;i>=0;i--)

#define peR(i,n) for(i=n;i>0;i--)

#define mem(a,b) memset(a,b,sizeof(a))

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

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

#define pi 3.1415926535897932384626433832795

#define vi vector<long long>

#define vs vector<string>

#define vp(vi,x) cin>>x; vi.pb(x);

#define srch(a,x) binary_search(all(a),x)

#define LB(a,x) (lower_bound(all(a),x)-a.B)

#define UB(a,x) (upper_bound(all(a),x)-a.B)

typedef long long LL;

typedef unsigned long long ULL;

bool isprm(LL a){for(LL i=2;i*i<=a;i++){if(a%i==0)return false;}return true;}

bool palin(string a){for(int i=0;i<a.sz;i++){if(a.at(i)!=a.at(a.sz-i-1))return 0;} return 1;}

bool isVowel(char ch){ ch=toupper(ch); if(ch=='A'||ch=='U'||ch=='I'||ch=='O'||ch=='E') return true; return false;}

int toInt(string s){int sm;stringstream ss(s);ss>>sm;return sm;}

template<typename T>inline T gcd(T a,T b){if(a<0)return gcd(-a,b);if(b<0)return gcd(a,-b);return (b==0)?a:gcd(b,a%b);}

template<typename T>inline T lcm(T a,T b) {if(a<0)return lcm(-a,b);if(b<0)return lcm(a,-b);return a*(b/gcd(a,b));}

int main()

{

//    fast;

    LL i,j,k,n,m,l,s=0,x,y,tc=1,c=0;

    while(cin>>n)

    {

        vi a;

        rep(i,n)

        {

            vp(a,x);

        }

        s=0;

        c=1;

        rep(i,n)

        {

            if(a[i]==c)

            {

                c++;

            }

            else s++;

        }

        if(s==n) cout<<-1<<en;

        else

        cout<<s<<en;

    }

}




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



int main()

{

	int n; scanf("%d", &n);

	int now = 0;

	

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

		int inp; scanf("%d", &inp);

		if(inp == now + 1) now++;

	}

	if(now == 0) printf("-1\n");

	else printf("%d\n", n - now);

}

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

using namespace std;



int main( )

{

    int n, i, j, k = 1;

    cin >> n;

    for (i = 1; i <= n; i++) {

        cin >> j;

        if (j == k) {

            k++;

        }

    }

    k--;

    if (k == 0) {

        cout << "-1\n";

    }

    else {

        cout << (n - k) << '\n';

    }

    return 0;

}


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



int main()

{

	int n; scanf("%d", &n);

	int now = 0;

	

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

		int inp; scanf("%d", &inp);

		if(inp == now + 1) now++;

	}

	if(now == 0) printf("-1\n");

	else printf("%d\n", n - now);

}

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

using namespace std;

#define ll long long

int main()

{

ll a,b,c,d=0,e=1,i,j,k,l;

 cin>>a;

  vector<ll>v;

  for(i=1;i<=a;i++)

  {

  cin>>b;

  v.push_back(b);

  }

  for(i=0;i<a;i++)

  {

  if(v[i]==e){e++;d++;}

  }

 if(d==0)cout<<"-1\n";

 else {

   cout<<a-d<<endl;

      }

}

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



int main()

{

	int n; scanf("%d", &n);

	int now = 0;

	

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

		int inp; scanf("%d", &inp);

		if(inp == now + 1) now++;

	}

	if(now == 0) printf("-1\n");

	else printf("%d\n", n - now);

}

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



using namespace std;

typedef long long ll ;

const int N =  1e5 ;

int main()

{

    ll n ;

    cin >> n ;

    int arr[n];

    int x=0 ;

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

    {

        cin >> arr[i] ;

        if(arr[i]==x+1)

        {

            x++;

        }

    }

    if(x==0)

    {

        cout << -1 ;

    }

    else cout <<n-x;

    return 0;

}


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



int main()

{

	int n; scanf("%d", &n);

	int now = 0;

	

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

		int inp; scanf("%d", &inp);

		if(inp == now + 1) now++;

	}

	if(now == 0) printf("-1\n");

	else printf("%d\n", n - now);

}

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

#define ll long long



 

using namespace std;

int main()

{

    ll n, temp;

    cin>>n;

    ll j=1,ans=0;

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

    {

        cin>>temp;

        if(temp==j)

        {   j++;

            ans++;

        }

    }

    if(ans==0)

        cout<<"-1\n";

    else

        cout<<n-ans<<"\n";

    

}

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



int main()

{

	int n; scanf("%d", &n);

	int now = 0;

	

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

		int inp; scanf("%d", &inp);

		if(inp == now + 1) now++;

	}

	if(now == 0) printf("-1\n");

	else printf("%d\n", n - now);

}

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

#define IO ios::sync_with_stdio(0), cin.tie(0) 

using namespace std;

typedef long long ll;

const int maxn = 1e5 + 5;

const int inf = ~0u >> 1;

int main() {

    IO;

    int n;

    cin >> n;

    if (n % 10 == 7 || n / 10 % 10 == 7 || n / 100 == 7) {

        cout << "Yes" << endl;

    } 

    else

    {

        cout << "No" << endl;

    }

    

    return 0;

}

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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		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 s;

	while(cin>>s){

		bool f=0;

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

			if(s[i]=='7')	f=1;

			if(f) puts("Yes");

			else puts("No");

	}

	return 0;

}

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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		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;

#define elif else if

typedef long long ll;

typedef unsigned long long ull;

#define DEBUG(x) cout<<(x)<<endl

#define TGET(a,b) get<b>(a)



int main()

{

    int n;

    scanf("%d", &n);

    int flag = 0;

    while(n)

    {

        if(n%10==7)

        {

            flag = 1;

        }

        n /= 10;

    }

    if(flag)

    {

        printf("Yes\n");

    }

    else

    {

        printf("No\n");

    }

    

    return 0;

}


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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		printf("Yes\n");

	}else{

		printf("No\n");

	}

	return 0;

}

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

#include<cmath>

#include<iostream>

#include<algorithm>

#include<cstring>

#include<cstdlib>

#include<queue>

#include<map>

#include<stack>

#include<set>

#include<vector>

#include<functional>

using namespace std;

typedef long long LL;

bool flag=0;

inline LL read()

{

	LL kk=0,f=1;

	char cc=getchar();

	while(cc<'0'||cc>'9'){if(cc=='-')f=-1;cc=getchar();}

	while(cc>='0'&&cc<='9'){if(cc-'0'==7)flag=1;kk=(kk<<1)+(kk<<3)+cc-'0';cc=getchar();}

	return kk*f;

}

int main()

{

	int k=read();

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

	else printf("No\n");

}

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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		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 s;

	cin>>s;

	for(int i=0;i<s.length();i++)

	{

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

		{

			puts("Yes");

			return 0;

		}

	}

	puts("No");

 } 

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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		printf("Yes\n");

	}else{

		printf("No\n");

	}

	return 0;

}

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



using namespace std;



int main(){

    char c[5]={0};

    cin>>c;

    int T=5,f=0;

    while(T--){

        if(c[T]=='7'){

            f=1;

            break;

        }

    }

    if(f) cout<<"Yes";

    else cout<<"No";

    return 0;

}

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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		printf("Yes\n");

	}else{

		printf("No\n");

	}

	return 0;

}

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

#define pb push_back

#define mp make_pair

#define F(i,n) for (int i = 0; i < n; ++i)

#define F1(i,n) for (int i = 1; i <= n; ++i)

#define rof(i, n) for (int i = ((n) - 1); i >= 0; i--)

typedef long long ll;

using namespace std;

vector<int>ivec;

vector<long long>lvec;

vector< pair<int, int> > pvec;

set< int, greater<int> >gset;

set<int, less<int> >lset;

set< int, greater<int> > ::iterator s;

void setprint()

 {

     set<int>so;

     set<int> :: iterator s;

     for(s=so.begin();s!=so.end();s++)

        cout<<*s<<" ";

 }



int main()

{

    ios_base::sync_with_stdio(0);

    cin.tie(0);

    /*

    {

       Md: Saroar Zahan Sojib

       Green University of Bangladesh

       https://web.facebook.com/sojib.bdb

    }

   */



  int n;

  cin>>n;

  vector<int>vec;

  int reminder;

  while(n!=0)

  {

      reminder=n%10;

      n=n/10;

      vec.push_back(reminder);

  }

  int count=0;

  for(auto x:vec)

  {

      if(x==7)

      {

          count++;

      }

  }

  if(count==0)cout<<"No"<<endl;

  else

    cout<<"Yes"<<endl;



}

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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		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 num;

    cin>>num;



    while(num != 0){

        int temp = num%10;

        if(temp == 7){

            cout<<"Yes"<<endl;

            return 0;

        }

        num /= 10;

    }

    cout<<"No"<<endl;

    return 0;

}


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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		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 flg=0;



    while(n!=0)

    {

        if(n%10==7)

        {

            flg=1;

            cout << "Yes" << endl;

            break;



        }



        n=n/10;



    }



    if(flg==0)

    {

        cout << "No" <<endl;

    }

    return 0;

}


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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		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, fact = 1;



    scanf("%d", &n);



    int t = n;



    while(t) {

        t = t / 10;

        fact = fact * 10;

    }



    while(fact > 1) {

        fact = fact / 10;

        if(n / fact == 7) {

            printf("Yes\n");

            return 0;

        }

        n = n % fact;

    }

    printf("No\n");



    return 0;

}

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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		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;



#define ll long long

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



int main()

{

    fast_in_out;

    int n;

    cin >> n;

    bool found = false;

    while(n != 0)

    {

        if(n % 10 == 7)

        {

            found = true;

            break;

        }

        n /= 10;

    }

    if(found)

    {

        cout << "Yes";

    }

    else

    {

        cout << "No";

    }

}


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

#include<string.h>

int main(){

	

	char s[100];

	scanf("%s",s);

	int count=0;

	for(int i=0; i<strlen(s); i++){

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

			count++;

		}else {

			continue;

		}

	}

	if(count>=1){

		printf("Yes\n");

	}else{

		printf("No\n");

	}

	return 0;

}

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



using namespace std;



int main() {

string z,x;

cin>>x;

z="7";

int length,c;

length=x.size();

for(c=0;c<=length;c=c+1){

    if(z[0]==x[c]){

        cout<<"Yes";

        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 IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

using namespace std;

	const int maxn=55;

	int a[maxn][maxn];

	int n;

	int main(){

		IOS

		bool ans=1;

		string s;

		cin>>s;

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

			cout<<"Yes"<<endl;

		else

			cout<<"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<iostream>

using namespace std;

int main()

{

	int n,cnt=0;

	cin>>n;

	while(n)

	{

		if(n%10==7)

		   cnt++;

		n/=10;

	}

	if(cnt==0)

	    cout<<"No"<<endl;

	else

	    cout<<"Yes"<<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;



 int main()

 {

 	int n;

 	cin>>n;

 	if(n%10 == 7 || n/10%10 == 7 || n/100 == 7)

 	    cout<<"Yes"<<endl;

 	else cout<<"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 <iostream>

using namespace std;

int main()

{

	string n; 

	cin>>n;

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

	cout<<"Yes";

	else

	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>



#define forn(i, n) for (int i = 0; i < int(n); ++i)

#define for1(i, n) for (int i = 1; i <= int(n); ++i)

#define ms(a, x) memset(a, x, sizeof(a))

#define F first

#define S second

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



using namespace std;

typedef long long ll;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

mt19937 gen(random_device{}());



int main() {

    ios::sync_with_stdio(false);

    cin.tie(nullptr);

    string st;

    cin>>st;

    if(st.find('7')!=st.npos) cout<<"Yes";

    else cout<<"No";

    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;

int main()

{

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

    {

        char cc;

        cc=getchar();

        if(cc=='7'){cout<<"Yes"<<endl;return 0;}

    }

    cout<<"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;

int main(){

    int n;

    cin>>n;

    if(n%10==7||n/10%10==7||n/100==7){

        cout<<"Yes";

    }

    else 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()

{

         string a;

         cin>>a;

         int c=0,i;

         for(i=0;i<a.size();i++)

        {

           if(a[i]=='7') c++;

        }

 

         if(c>0) cout<<"Yes"<<endl;

         else cout<<"No"<<endl;

}

// 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;



    while (n)

    {

        if (n % 10 == 7)

        {

            cout << "Yes\n";

            return 0;

        }

        n/=10;

    }

    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;



#define     bip(a)         __builtin_popcount(a)

#define     ull            unsigned long long

#define     ll             long long

#define     u              unsigned

#define     pb             push_back

#define     pf             push_front ///for list

#define     mp             make_pair

#define     is             insert

#define     ff             first

#define     ss             second

#define     vi             vector<int>

#define     vl             vector<ll>

#define     si             set<int>

#define     sl             set<ll>

#define     li             list<int>

#define     pii            pair<int,int>

#define     pll            pair<ll,ll>

#define     mii            map<int,int>

#define     mll            map<ll,ll>

#define     ma             INT_MAX

#define     mi             INT_MIN

#define     mod            1000000007

#define     pi             3.14159265359

#define     e              2.71828182846

#define     inf            1000000000000000LL

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

#define     lb(a,b,c)      lower_bound(a,a+b,c)-a

#define     ub(a,b,c)      upper_bound(a,a+b,c)-a

#define     lbv(a,c)       lower_bound(all(a),c)-a.begin()

#define     ubv(a,c)       upper_bound(all(a),c)-a.begin()

#define     srt1(a,b)      sort(a,a+b)

#define     srt2(a,b)      sort(a,a+b,greater<int>())

#define     gcd(a,b)       __gcd(a,b)

#define     lcm(a,b)       (a*(b/gcd(a,b)))

#define     harmonic(n)    0.57721566490153286l+log(n)

#define     mem(a, b)      memset(a, b, sizeof(a))



///s.lower_bound(a),s.upper_bound(a), for set

///nCr(x, y) = nCr(x-1, y-1) + nCr(x-1, y)

///factorial digit   ((n * log10(n / e) +log10(2 * pi * n) /2.0));



int main()

{

	/*int n,q,l,r,ac=0;

	string str;

	cin>>n>>q>>str;

	int pre[n+5];

	mem(pre,0);

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

		if(str[i]=='A'&&str[i+1]=='C'){

			ac++;

			pre[i+2]=ac;

			pre[i+1]=ac-1;

			i+=2;

		}

		else{

			pre[i+1]=ac;

			i++;

		}

	}*/

	/*for(int i=1;i<=n;i++)cout<<pre[i]<<" ";

	cout<<endl;*/

	/*while(q--){

		cin>>l>>r;

		cout<<pre[r]-pre[l]<<endl;

	}*/

	/*char ch;

	cin>>ch;

	if(ch=='A')cout<<"T\n";

	if(ch=='T')cout<<"A\n";

	if(ch=='G')cout<<"C\n";

	if(ch=='C')cout<<"G\n";*/

	string str;

	cin>>str;

	if(count(str.begin(),str.end(),'7')>0)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 <iostream>

using namespace std;

int main()

{

int n,a,b,c;

cin>>n;

a=n/100;

b=n/10%10;

c=n%10;

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

else if (a==7 and b==7 and c==7)cout<<"Yes"<<endl;

else if (a==7 and b==7)cout<<"Yes"<<endl;

else if (a==7 and c==7)cout<<"Yes"<<endl;

else if (b==7 and c==7)cout<<"Yes"<<endl;

else cout<<"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: