stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

#define ll  long long

struct triple{ll   x=0;ll   y=0;ll   z=0;void mt(ll   a,ll   b, ll   c){x=a;y=b;z=c;} void holify(){x=69;y=420;z=911;}};

#define pb push_back

#define x first 

#define y second

#define vec vector

#define pf(vec,val) vec.insert(vec.begin(),val)

#define mp make_pair

#define pii pair<ll  ,ll  >

#define maxll  69696991169

#define minll  -69696942069

#define sz(vec) (ll )vec.size()

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

#define loop(mep) for(auto it = mep.begin();it!=mep.end();++it)

//#define usaco_crap(str)ifstream cin((string)str + (string)".in"); ofstream cout ((string)str+(string)".out");

#define range(n) =0;i<(n);i++

#define prll  (vec) for(auto it=vec.begin();it!=vec.end();++it)cerr<<*it<<" ";cerr<<endl;

// mt make triple // triple.mt(a,b,c) //

int  main(){

	int n; cin>>n;

	int flag = 0;

	for(int i = 1; i < 10 ; i++){

		for(int j = 1 ; j< 10 ; j++){

			if(i*j==n)flag=1;

		}

	}

	if(flag)cout<<"Yes";

	else cout<<"No";

}


// Faster C++ solution for the same problem (measured speedup: 5.3169902535061x):
#include<stdio.h>

	int main(){

		int a;

		scanf("%d",&a);

		

		for (int i =1 ;i<=9 ; i++){

			for(int j=1;j<=9;j++){

				if(i*j==a){

					puts("Yes");

					return 0;

				}

			}

		}



	puts("No");

}
// Slower C++ solution:
#include <iostream>

#include <math.h>



using namespace std;



int main()

{

    int N; cin >> N;

    

    for (double i = 1; i <= 9; i++) {

        if ((N / i) == floor(N/i) and (N / i) >= 1 and (N / i) <= 9) {

            cout << "Yes";

            return 0;

        }

    }



    cout << "No";

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.321054908085674x):
#include<stdio.h>

	int main(){

		int a;

		scanf("%d",&a);

		

		for (int i =1 ;i<=9 ; i++){

			for(int j=1;j<=9;j++){

				if(i*j==a){

					puts("Yes");

					return 0;

				}

			}

		}



	puts("No");

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define ll long long

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

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

#define ff first

#define ss second

#define p pair<int,int>

#define pb push_back

#define endl '\n'

#define w(t) ll test;cin>>test;while(test--)

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

#define pi acos(-1)



int main()

{

    fast;

    int n;

    cin>>n;

    //bool flag=false;

    for(int i=1;i<=9;i++)

    {

        int temp;

        if(n%i==0)

        {

            temp=n/i;

       // cout<<i<<" "<<temp<<endl;

        if(temp<=9&&temp>=1)

        {

            cout<<"Yes"<<endl;

            return 0;

        }



        }

    }

    cout<<"No"<<endl;

}




// Faster C++ solution for the same problem (measured speedup: 5.390795809627206x):
#include<stdio.h>

	int main(){

		int a;

		scanf("%d",&a);

		

		for (int i =1 ;i<=9 ; i++){

			for(int j=1;j<=9;j++){

				if(i*j==a){

					puts("Yes");

					return 0;

				}

			}

		}



	puts("No");

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define lli long long int

#define pii pair<int,int>

#define pll pair<lli,lli>

#define vi  vector<int>

#define vii vector<pair<int,int>>

#define vll vector<lli>

#define pb  push_back

#define mp  make_pair

#define ss  second

#define ff  first

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

const lli mod = 1e9+7;

//#ifndef ONLINE_JUDGE

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

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

//    #endif

lli p2(lli n){lli ans=0;while(n>=1){n=n/2;ans++;}return ans;}

lli b_ex(lli n,lli a){lli res=1;while(n){if(n%2==1)res*=a;n>>=1;a=a*a;}return res;}

lli ncr(lli n,lli k){lli ans=1;if(k>n-k)k=n-k;for(lli i=1;i<=k;i++){ans*=(n-i+1),ans=ans/i;}return ans;}

lli add(lli a,lli b){return(a+b)%mod;}

lli mul(lli a,lli b){return (a*b)%mod;}

lli power(lli a , lli n){lli ans = 1;while(n){if(n % 2)ans = (ans * a) % mod;n /= 2;a = (a * a) % mod;}return ans;}

lli lcm(lli a,lli b)

{

    lli ans=(a*b)/(__gcd(a,b));

    return ans;

}



int main()

{

    int n;cin>>n;

    if(n<=0 or n>81)cout<<"No\n";

    else

    {

        for(int i=1;i<=9;i++)

        {

            if(n%i==0 and n/i <= 9)

            {

                cout<<"Yes\n";return 0;

            }

        }

        cout<<"No\n";

    }

}








// Faster C++ solution for the same problem (measured speedup: 5.274791674387041x):
#include<stdio.h>

	int main(){

		int a;

		scanf("%d",&a);

		

		for (int i =1 ;i<=9 ; i++){

			for(int j=1;j<=9;j++){

				if(i*j==a){

					puts("Yes");

					return 0;

				}

			}

		}



	puts("No");

}
// Slower C++ solution:
//Bismillahir Rahmanir Rahim

//In the name of Allah the Most Merciful

//Astagfirullahil laji Rabbi Minkulli Jamio Watubi elai

#include<bits/stdc++.h>

using namespace std;

#define ll long long int

#define double long double

#define MX 10000008

#define cs int t;cin>>t;while(t--)

int i,j,k;

int r,s;

int main()

 {

    int x;

    cin>>x;

    int p=0;

    for(i=1;i<=9;i++)

    {

        if(x%i==0)

        {

            int k=x/i;

            if(k>=1&&k<=9)

            {

                p=k;



            }

        }

    }

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

    else cout<<"No"<<endl;

 }


// Faster C++ solution for the same problem (measured speedup: 5.332016853828078x):
#include<stdio.h>

	int main(){

		int a;

		scanf("%d",&a);

		

		for (int i =1 ;i<=9 ; i++){

			for(int j=1;j<=9;j++){

				if(i*j==a){

					puts("Yes");

					return 0;

				}

			}

		}



	puts("No");

}
// Slower C++ solution:
//

// author  : Abhishek

//

#include <bits/stdc++.h>



#define ll long long

#define db double

#define deb(x) cout<<#x<<": "<<x<<"\n"

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

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

#define Fo(i, k, n) for(i=k ; i<n ; i++)



using namespace std;



const ll mod = 1000000007;

const ll N = 300005;  // array size

const ll inf = 10000000000000000;





void print(vector<ll> v) {

    ll i;

    fo(i, v.size())cout << v[i] << " ";

    cout << "\n";

}



vector<ll> read(ll n) {

    ll i;

    vector<ll> arr(n, 0);

    fo(i, n)cin >> arr[i];

    return arr;

}



vector<ll> read(ll n, ll k) {

    ll i;

    vector<ll> arr(n + k, 0);

    Fo(i, k, n + k)cin >> arr[i];

    return arr;

}



bool isEven(ll n) {

    return n % 2 == 0;

}



bool isOdd(ll n) {

    return n % 2 != 0;

}





void solve() {

    ll t;

    cin >> (t);

    while (t--) {



        ll i, j, k, n, m, p, q, x, y, a, b, l, r;





    }

}



int main() {



//#ifndef ONLINE_JUDGE

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

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

//#endif



    ios_base::sync_with_stdio(false);

    cin.tie(NULL);



    ll i, j, k, n, m, p, q, x, y, a, b, l, r;





    cin>>n;



    Fo(i,1,10){

        Fo(j,1,10){

            if(i*j == n){

                cout<<"Yes\n";

                return 0;

            }

        }

    }



    cout<<"No\n";



    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.435677534464118x):
#include<stdio.h>

	int main(){

		int a;

		scanf("%d",&a);

		

		for (int i =1 ;i<=9 ; i++){

			for(int j=1;j<=9;j++){

				if(i*j==a){

					puts("Yes");

					return 0;

				}

			}

		}



	puts("No");

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

#define ll  long long

struct triple{ll   x=0;ll   y=0;ll   z=0;void mt(ll   a,ll   b, ll   c){x=a;y=b;z=c;} void holify(){x=69;y=420;z=911;}};

#define pb push_back

#define x first 

#define y second

#define vec vector

#define pf(vec,val) vec.insert(vec.begin(),val)

#define mp make_pair

#define pii pair<ll  ,ll  >

#define maxll  69696991169

#define minll  -69696942069

#define sz(vec) (ll )vec.size()

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

#define loop(mep) for(auto it = mep.begin();it!=mep.end();++it)

//#define usaco_crap(str)ifstream cin((string)str + (string)".in"); ofstream cout ((string)str+(string)".out");

#define range(n) =0;i<(n);i++

#define prll  (vec) for(auto it=vec.begin();it!=vec.end();++it)cerr<<*it<<" ";cerr<<endl;

// mt make triple // triple.mt(a,b,c) //

int  main(){

	int n; cin>>n;

	int flag = 0;

	for(int i = 1; i < 10 ; i++){

		for(int j = 1 ; j< 10 ; j++){

			if(i*j==n)flag=1;

		}

	}

	if(flag)cout<<"Yes";

	else cout<<"No";

}


// Faster C++ solution for the same problem (measured speedup: 5.318083054101105x):
#include <stdio.h>

 

int main(){

	

int n;

scanf ("%d", &n);

 

int bilangan = 0;

 // harus dibawah 10 

for (int i = 1; i < 10; i++){ // bilanagn yang pertama

	for (int j = 1; j < 10; j++){ // bilangan yang ke -2 

		

		if (n == i * j){

			bilangan = 1;

			break;

		}

}

/* If N can be represented as the product of two integers between 1 and 9 (inclusive), print Yes; if it cannot, print No.

 */

	}

	if (bilangan == 1){

		puts ("Yes");

	}

	else if (bilangan == 0) {

		puts ("No");

	}

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include <math.h>



using namespace std;



int main()

{

    int N; cin >> N;

    

    for (double i = 1; i <= 9; i++) {

        if ((N / i) == floor(N/i) and (N / i) >= 1 and (N / i) <= 9) {

            cout << "Yes";

            return 0;

        }

    }



    cout << "No";

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.3221485440888205x):
#include <stdio.h>

 

int main(){

	

int n;

scanf ("%d", &n);

 

int bilangan = 0;

 // harus dibawah 10 

for (int i = 1; i < 10; i++){ // bilanagn yang pertama

	for (int j = 1; j < 10; j++){ // bilangan yang ke -2 

		

		if (n == i * j){

			bilangan = 1;

			break;

		}

}

/* If N can be represented as the product of two integers between 1 and 9 (inclusive), print Yes; if it cannot, print No.

 */

	}

	if (bilangan == 1){

		puts ("Yes");

	}

	else if (bilangan == 0) {

		puts ("No");

	}

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define ll long long

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

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

#define ff first

#define ss second

#define p pair<int,int>

#define pb push_back

#define endl '\n'

#define w(t) ll test;cin>>test;while(test--)

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

#define pi acos(-1)



int main()

{

    fast;

    int n;

    cin>>n;

    //bool flag=false;

    for(int i=1;i<=9;i++)

    {

        int temp;

        if(n%i==0)

        {

            temp=n/i;

       // cout<<i<<" "<<temp<<endl;

        if(temp<=9&&temp>=1)

        {

            cout<<"Yes"<<endl;

            return 0;

        }



        }

    }

    cout<<"No"<<endl;

}




// Faster C++ solution for the same problem (measured speedup: 5.391903779472444x):
#include <stdio.h>

 

int main(){

	

int n;

scanf ("%d", &n);

 

int bilangan = 0;

 // harus dibawah 10 

for (int i = 1; i < 10; i++){ // bilanagn yang pertama

	for (int j = 1; j < 10; j++){ // bilangan yang ke -2 

		

		if (n == i * j){

			bilangan = 1;

			break;

		}

}

/* If N can be represented as the product of two integers between 1 and 9 (inclusive), print Yes; if it cannot, print No.

 */

	}

	if (bilangan == 1){

		puts ("Yes");

	}

	else if (bilangan == 0) {

		puts ("No");

	}

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define lli long long int

#define pii pair<int,int>

#define pll pair<lli,lli>

#define vi  vector<int>

#define vii vector<pair<int,int>>

#define vll vector<lli>

#define pb  push_back

#define mp  make_pair

#define ss  second

#define ff  first

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

const lli mod = 1e9+7;

//#ifndef ONLINE_JUDGE

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

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

//    #endif

lli p2(lli n){lli ans=0;while(n>=1){n=n/2;ans++;}return ans;}

lli b_ex(lli n,lli a){lli res=1;while(n){if(n%2==1)res*=a;n>>=1;a=a*a;}return res;}

lli ncr(lli n,lli k){lli ans=1;if(k>n-k)k=n-k;for(lli i=1;i<=k;i++){ans*=(n-i+1),ans=ans/i;}return ans;}

lli add(lli a,lli b){return(a+b)%mod;}

lli mul(lli a,lli b){return (a*b)%mod;}

lli power(lli a , lli n){lli ans = 1;while(n){if(n % 2)ans = (ans * a) % mod;n /= 2;a = (a * a) % mod;}return ans;}

lli lcm(lli a,lli b)

{

    lli ans=(a*b)/(__gcd(a,b));

    return ans;

}



int main()

{

    int n;cin>>n;

    if(n<=0 or n>81)cout<<"No\n";

    else

    {

        for(int i=1;i<=9;i++)

        {

            if(n%i==0 and n/i <= 9)

            {

                cout<<"Yes\n";return 0;

            }

        }

        cout<<"No\n";

    }

}








// Faster C++ solution for the same problem (measured speedup: 5.275875801911348x):
#include <stdio.h>

 

int main(){

	

int n;

scanf ("%d", &n);

 

int bilangan = 0;

 // harus dibawah 10 

for (int i = 1; i < 10; i++){ // bilanagn yang pertama

	for (int j = 1; j < 10; j++){ // bilangan yang ke -2 

		

		if (n == i * j){

			bilangan = 1;

			break;

		}

}

/* If N can be represented as the product of two integers between 1 and 9 (inclusive), print Yes; if it cannot, print No.

 */

	}

	if (bilangan == 1){

		puts ("Yes");

	}

	else if (bilangan == 0) {

		puts ("No");

	}

	return 0;

}
// Slower C++ solution:
//Bismillahir Rahmanir Rahim

//In the name of Allah the Most Merciful

//Astagfirullahil laji Rabbi Minkulli Jamio Watubi elai

#include<bits/stdc++.h>

using namespace std;

#define ll long long int

#define double long double

#define MX 10000008

#define cs int t;cin>>t;while(t--)

int i,j,k;

int r,s;

int main()

 {

    int x;

    cin>>x;

    int p=0;

    for(i=1;i<=9;i++)

    {

        if(x%i==0)

        {

            int k=x/i;

            if(k>=1&&k<=9)

            {

                p=k;



            }

        }

    }

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

    else cout<<"No"<<endl;

 }


// Faster C++ solution for the same problem (measured speedup: 5.3331127428390825x):
#include <stdio.h>

 

int main(){

	

int n;

scanf ("%d", &n);

 

int bilangan = 0;

 // harus dibawah 10 

for (int i = 1; i < 10; i++){ // bilanagn yang pertama

	for (int j = 1; j < 10; j++){ // bilangan yang ke -2 

		

		if (n == i * j){

			bilangan = 1;

			break;

		}

}

/* If N can be represented as the product of two integers between 1 and 9 (inclusive), print Yes; if it cannot, print No.

 */

	}

	if (bilangan == 1){

		puts ("Yes");

	}

	else if (bilangan == 0) {

		puts ("No");

	}

	return 0;

}
// Slower C++ solution:
//

// author  : Abhishek

//

#include <bits/stdc++.h>



#define ll long long

#define db double

#define deb(x) cout<<#x<<": "<<x<<"\n"

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

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

#define Fo(i, k, n) for(i=k ; i<n ; i++)



using namespace std;



const ll mod = 1000000007;

const ll N = 300005;  // array size

const ll inf = 10000000000000000;





void print(vector<ll> v) {

    ll i;

    fo(i, v.size())cout << v[i] << " ";

    cout << "\n";

}



vector<ll> read(ll n) {

    ll i;

    vector<ll> arr(n, 0);

    fo(i, n)cin >> arr[i];

    return arr;

}



vector<ll> read(ll n, ll k) {

    ll i;

    vector<ll> arr(n + k, 0);

    Fo(i, k, n + k)cin >> arr[i];

    return arr;

}



bool isEven(ll n) {

    return n % 2 == 0;

}



bool isOdd(ll n) {

    return n % 2 != 0;

}





void solve() {

    ll t;

    cin >> (t);

    while (t--) {



        ll i, j, k, n, m, p, q, x, y, a, b, l, r;





    }

}



int main() {



//#ifndef ONLINE_JUDGE

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

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

//#endif



    ios_base::sync_with_stdio(false);

    cin.tie(NULL);



    ll i, j, k, n, m, p, q, x, y, a, b, l, r;





    cin>>n;



    Fo(i,1,10){

        Fo(j,1,10){

            if(i*j == n){

                cout<<"Yes\n";

                return 0;

            }

        }

    }



    cout<<"No\n";



    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.436794728846786x):
#include <stdio.h>

 

int main(){

	

int n;

scanf ("%d", &n);

 

int bilangan = 0;

 // harus dibawah 10 

for (int i = 1; i < 10; i++){ // bilanagn yang pertama

	for (int j = 1; j < 10; j++){ // bilangan yang ke -2 

		

		if (n == i * j){

			bilangan = 1;

			break;

		}

}

/* If N can be represented as the product of two integers between 1 and 9 (inclusive), print Yes; if it cannot, print No.

 */

	}

	if (bilangan == 1){

		puts ("Yes");

	}

	else if (bilangan == 0) {

		puts ("No");

	}

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

#define ll  long long

struct triple{ll   x=0;ll   y=0;ll   z=0;void mt(ll   a,ll   b, ll   c){x=a;y=b;z=c;} void holify(){x=69;y=420;z=911;}};

#define pb push_back

#define x first 

#define y second

#define vec vector

#define pf(vec,val) vec.insert(vec.begin(),val)

#define mp make_pair

#define pii pair<ll  ,ll  >

#define maxll  69696991169

#define minll  -69696942069

#define sz(vec) (ll )vec.size()

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

#define loop(mep) for(auto it = mep.begin();it!=mep.end();++it)

//#define usaco_crap(str)ifstream cin((string)str + (string)".in"); ofstream cout ((string)str+(string)".out");

#define range(n) =0;i<(n);i++

#define prll  (vec) for(auto it=vec.begin();it!=vec.end();++it)cerr<<*it<<" ";cerr<<endl;

// mt make triple // triple.mt(a,b,c) //

int  main(){

	int n; cin>>n;

	int flag = 0;

	for(int i = 1; i < 10 ; i++){

		for(int j = 1 ; j< 10 ; j++){

			if(i*j==n)flag=1;

		}

	}

	if(flag)cout<<"Yes";

	else cout<<"No";

}


// Faster C++ solution for the same problem (measured speedup: 5.315586295351271x):
#include <stdio.h>



int main(){

	int angka;

	scanf("%d", &angka);

	int count = 0;

	for(int i = 1; i <= 9; i++){

		for(int j = 1; j <= 9; j++){

			if(angka == i * j){

				count++;

			}

		}

	}

	if(count != 0){

		puts("Yes");

	} else{

		puts("No");

	}

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include <math.h>



using namespace std;



int main()

{

    int N; cin >> N;

    

    for (double i = 1; i <= 9; i++) {

        if ((N / i) == floor(N/i) and (N / i) >= 1 and (N / i) <= 9) {

            cout << "Yes";

            return 0;

        }

    }



    cout << "No";

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.319649876653545x):
#include <stdio.h>



int main(){

	int angka;

	scanf("%d", &angka);

	int count = 0;

	for(int i = 1; i <= 9; i++){

		for(int j = 1; j <= 9; j++){

			if(angka == i * j){

				count++;

			}

		}

	}

	if(count != 0){

		puts("Yes");

	} else{

		puts("No");

	}

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define ll long long

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

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

#define ff first

#define ss second

#define p pair<int,int>

#define pb push_back

#define endl '\n'

#define w(t) ll test;cin>>test;while(test--)

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

#define pi acos(-1)



int main()

{

    fast;

    int n;

    cin>>n;

    //bool flag=false;

    for(int i=1;i<=9;i++)

    {

        int temp;

        if(n%i==0)

        {

            temp=n/i;

       // cout<<i<<" "<<temp<<endl;

        if(temp<=9&&temp>=1)

        {

            cout<<"Yes"<<endl;

            return 0;

        }



        }

    }

    cout<<"No"<<endl;

}




// Faster C++ solution for the same problem (measured speedup: 5.389372363019803x):
#include <stdio.h>



int main(){

	int angka;

	scanf("%d", &angka);

	int count = 0;

	for(int i = 1; i <= 9; i++){

		for(int j = 1; j <= 9; j++){

			if(angka == i * j){

				count++;

			}

		}

	}

	if(count != 0){

		puts("Yes");

	} else{

		puts("No");

	}

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define lli long long int

#define pii pair<int,int>

#define pll pair<lli,lli>

#define vi  vector<int>

#define vii vector<pair<int,int>>

#define vll vector<lli>

#define pb  push_back

#define mp  make_pair

#define ss  second

#define ff  first

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

const lli mod = 1e9+7;

//#ifndef ONLINE_JUDGE

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

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

//    #endif

lli p2(lli n){lli ans=0;while(n>=1){n=n/2;ans++;}return ans;}

lli b_ex(lli n,lli a){lli res=1;while(n){if(n%2==1)res*=a;n>>=1;a=a*a;}return res;}

lli ncr(lli n,lli k){lli ans=1;if(k>n-k)k=n-k;for(lli i=1;i<=k;i++){ans*=(n-i+1),ans=ans/i;}return ans;}

lli add(lli a,lli b){return(a+b)%mod;}

lli mul(lli a,lli b){return (a*b)%mod;}

lli power(lli a , lli n){lli ans = 1;while(n){if(n % 2)ans = (ans * a) % mod;n /= 2;a = (a * a) % mod;}return ans;}

lli lcm(lli a,lli b)

{

    lli ans=(a*b)/(__gcd(a,b));

    return ans;

}



int main()

{

    int n;cin>>n;

    if(n<=0 or n>81)cout<<"No\n";

    else

    {

        for(int i=1;i<=9;i++)

        {

            if(n%i==0 and n/i <= 9)

            {

                cout<<"Yes\n";return 0;

            }

        }

        cout<<"No\n";

    }

}








// Faster C++ solution for the same problem (measured speedup: 5.273398858821619x):
#include <stdio.h>



int main(){

	int angka;

	scanf("%d", &angka);

	int count = 0;

	for(int i = 1; i <= 9; i++){

		for(int j = 1; j <= 9; j++){

			if(angka == i * j){

				count++;

			}

		}

	}

	if(count != 0){

		puts("Yes");

	} else{

		puts("No");

	}

	return 0;

}
// Slower C++ solution:
//Bismillahir Rahmanir Rahim

//In the name of Allah the Most Merciful

//Astagfirullahil laji Rabbi Minkulli Jamio Watubi elai

#include<bits/stdc++.h>

using namespace std;

#define ll long long int

#define double long double

#define MX 10000008

#define cs int t;cin>>t;while(t--)

int i,j,k;

int r,s;

int main()

 {

    int x;

    cin>>x;

    int p=0;

    for(i=1;i<=9;i++)

    {

        if(x%i==0)

        {

            int k=x/i;

            if(k>=1&&k<=9)

            {

                p=k;



            }

        }

    }

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

    else cout<<"No"<<endl;

 }


// Faster C++ solution for the same problem (measured speedup: 5.330608927879995x):
#include <stdio.h>



int main(){

	int angka;

	scanf("%d", &angka);

	int count = 0;

	for(int i = 1; i <= 9; i++){

		for(int j = 1; j <= 9; j++){

			if(angka == i * j){

				count++;

			}

		}

	}

	if(count != 0){

		puts("Yes");

	} else{

		puts("No");

	}

	return 0;

}