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

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

#define debug  freopen("in.txt","r",stdin),freopen("out.txt","w",stdout);

#define pb push_back

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

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int maxn = 1e6+10;

const int maxM = 1e6+10;

const int inf = 0x3f3f3f3f;



int K,A,B;

int main(){

	// debug;

	ios;



	cin>>K>>A>>B;

	if(B/K - (A-1)/K > 0) cout<<"OK\n";

	else cout<<"NG\n";



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.371976455650526x):
#include <stdio.h>



int main() {



	int K, A, B;

	

	scanf("%d", &K);

	scanf("%d %d", &A, &B);

	

	int M = B%K;

	int N = B-M;

	

	if (N>=A) {

		printf("OK\n");

	}

	else {

		printf("NG\n");

	}



    return 0;

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

using namespace std;

int k,a,b,gg;

int main(){

	cin>>k>>a>>b;

	for (int i=1;i<=1000;i++){

		gg=k*i;

		if (a<=gg && gg<=b){

			cout<<"OK"<<endl;

			return 0;

		}

	}

	cout<<"NG"<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.308896255894674x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

using namespace std;

int k,a,b;

int main(){

	cin>>k>>a>>b;

	if(a/k<b/k || (a%k==0 || b%k==0)) cout<<"OK"<<endl;

	else cout<<"NG"<<endl;

	

} 

// Faster C++ solution for the same problem (measured speedup: 5.3089318831734635x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

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

#define for0(i, n) for(int i = 0; i < n; ++i)

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

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

/*=================================================================================*/



int k, a, b, flag;



int main() {

   cin >> k >> a >> b;

   for(int i = a; i <= b; ++i) {

      if(!(i%k)) flag = 1;

   }

   flag ? puts("OK") : puts("NG");

}

// Faster C++ solution for the same problem (measured speedup: 5.241624161860046x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

using namespace std;

long long k,a,b;

int main(){

	cin>>k>>a>>b;

	if(b>=a/k*k+k||a==a/k*k||b==a/k*k+k) cout<<"OK"<<endl;

	else cout<<"NG"<<endl;

	return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 5.309051360368024x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

using namespace std;

int main()

{

    int k,a,b;

    cin>>k>>a>>b;

    for(int i=a;i<=b;i++){

        if(i%k==0){cout<<"OK";return 0;}

    }cout<<"NG";

}


// Faster C++ solution for the same problem (measured speedup: 5.295686241363977x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

}
// Slower C++ solution:


#include<bits/stdc++.h>



using namespace std;



int main(){

    int n,i,j=0,k,l,r;

    cin>>k>>l>>r;

    for(i=l;i<=r;i++){

        if(i%k==0){cout<<"OK";j=1;break;}

    }

    if(j==0){cout<<"NG";}



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.295726143344354x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

#include<string>

#include<cstring>

using namespace std;

int main()

{

    int a,b,k;

    cin>>k;

    cin>>a>>b;

    int i;

    int j=0;

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

    {

    if(i%k==0)

    {

    j++;

    break;

    }

    }

    if(j)

    cout<<"OK"<<endl;

    else

    cout<<"NG"<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.309784421652104x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

using namespace std;

int main()

{

    int k,a,b,l=0;

    cin >> k;

    cin >> a>> b;

    for(int i=a; i<=b; i++)

    {

        if(i%k==0)

        {

            l++;

            cout<< "OK"<<endl;

            break;

        }

        else

            continue;

    }

    if(l==0)

        cout<< "NG"<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 5.309863210549941x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

using namespace std;

int main()

{

    int k,a,b,j,l,i;

    cin>>k>>a>>b;

    j=0;

    for(i=1;i>0;i++)

    {

        l=k*i;

        if(l>=a&&l<=b)

        {

            j=1;

        }

        if(l>b)

            break;

    }

    if(j==1)

        cout<<"OK";

    else

        cout<<"NG";



}


// Faster C++ solution for the same problem (measured speedup: 5.29448893890941x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

using namespace std;

int main()

{

    int div,a,b,count;

    while(cin>>div>>a>>b)

    {

        count=0;

        for(int i=a;i<=b;i++)

        {

            if(i%div==0)

            {

                count++;

                 break;

            }

        }

        if(count!=0)

        {

            cout<<"OK"<<endl;

        }

        else

        {

            cout<<"NG"<<endl;

        }

    }

}


// Faster C++ solution for the same problem (measured speedup: 5.31100977415522x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

#include <cmath>

using namespace std;

int main() {

    int k,a,b;

    cin>>k;

    cin>>a>>b;

    if(a%k==0||b%k==0){

    	cout<<"OK";

    }else if((((a/k)+1)*k)<b){

    	cout<<"OK";

    }else cout<<"NG";

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.295726157641015x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

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

#define debug  freopen("in.txt","r",stdin),freopen("out.txt","w",stdout);

#define pb push_back

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

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int maxn = 1e6+10;

const int maxM = 1e6+10;

const int inf = 0x3f3f3f3f;



int K,A,B;

int main(){

	// debug;

	ios;



	cin>>K>>A>>B;

	if(B/K - (A-1)/K > 0) cout<<"OK\n";

	else cout<<"NG\n";



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.370324075135072x):
#include <stdio.h>





int main(){

    int multi; // 5

    int multiAfter=0;

    int tbawah, tatas; // 14 16

    scanf("%d", &multi);

    scanf("%d %d", &tbawah, &tatas);

    while(multiAfter<=tatas){

        multiAfter=multiAfter+multi;

        if(multiAfter>=tbawah && multiAfter<=tatas){

            printf("OK\n");

            break;

        }

        else if(multiAfter>=tatas){

            printf("NG\n");

            break;

        }    

        else{

            continue;

        }

    }



    return 0;

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

using namespace std;

int k,a,b,gg;

int main(){

	cin>>k>>a>>b;

	for (int i=1;i<=1000;i++){

		gg=k*i;

		if (a<=gg && gg<=b){

			cout<<"OK"<<endl;

			return 0;

		}

	}

	cout<<"NG"<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.30866704933363x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

using namespace std;

int k,a,b;

int main(){

	cin>>k>>a>>b;

	if(a/k<b/k || (a%k==0 || b%k==0)) cout<<"OK"<<endl;

	else cout<<"NG"<<endl;

	

} 

// Faster C++ solution for the same problem (measured speedup: 5.3087026750742465x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

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

#define for0(i, n) for(int i = 0; i < n; ++i)

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

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

/*=================================================================================*/



int k, a, b, flag;



int main() {

   cin >> k >> a >> b;

   for(int i = a; i <= b; ++i) {

      if(!(i%k)) flag = 1;

   }

   flag ? puts("OK") : puts("NG");

}

// Faster C++ solution for the same problem (measured speedup: 5.241397859707863x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

using namespace std;

long long k,a,b;

int main(){

	cin>>k>>a>>b;

	if(b>=a/k*k+k||a==a/k*k||b==a/k*k+k) cout<<"OK"<<endl;

	else cout<<"NG"<<endl;

	return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 5.308822147110492x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

using namespace std;

int main()

{

    int k,a,b;

    cin>>k>>a>>b;

    for(int i=a;i<=b;i++){

        if(i%k==0){cout<<"OK";return 0;}

    }cout<<"NG";

}


// Faster C++ solution for the same problem (measured speedup: 5.295457605132783x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

}
// Slower C++ solution:


#include<bits/stdc++.h>



using namespace std;



int main(){

    int n,i,j=0,k,l,r;

    cin>>k>>l>>r;

    for(i=l;i<=r;i++){

        if(i%k==0){cout<<"OK";j=1;break;}

    }

    if(j==0){cout<<"NG";}



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.29549750539043x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

#include<string>

#include<cstring>

using namespace std;

int main()

{

    int a,b,k;

    cin>>k;

    cin>>a>>b;

    int i;

    int j=0;

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

    {

    if(i%k==0)

    {

    j++;

    break;

    }

    }

    if(j)

    cout<<"OK"<<endl;

    else

    cout<<"NG"<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.309555176745347x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

using namespace std;

int main()

{

    int k,a,b,l=0;

    cin >> k;

    cin >> a>> b;

    for(int i=a; i<=b; i++)

    {

        if(i%k==0)

        {

            l++;

            cout<< "OK"<<endl;

            break;

        }

        else

            continue;

    }

    if(l==0)

        cout<< "NG"<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 5.309633962241547x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

using namespace std;

int main()

{

    int k,a,b,j,l,i;

    cin>>k>>a>>b;

    j=0;

    for(i=1;i>0;i++)

    {

        l=k*i;

        if(l>=a&&l<=b)

        {

            j=1;

        }

        if(l>b)

            break;

    }

    if(j==1)

        cout<<"OK";

    else

        cout<<"NG";



}


// Faster C++ solution for the same problem (measured speedup: 5.294260354370614x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

using namespace std;

int main()

{

    int div,a,b,count;

    while(cin>>div>>a>>b)

    {

        count=0;

        for(int i=a;i<=b;i++)

        {

            if(i%div==0)

            {

                count++;

                 break;

            }

        }

        if(count!=0)

        {

            cout<<"OK"<<endl;

        }

        else

        {

            cout<<"NG"<<endl;

        }

    }

}


// Faster C++ solution for the same problem (measured speedup: 5.31078047634503x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

#include <cmath>

using namespace std;

int main() {

    int k,a,b;

    cin>>k;

    cin>>a>>b;

    if(a%k==0||b%k==0){

    	cout<<"OK";

    }else if((((a/k)+1)*k)<b){

    	cout<<"OK";

    }else cout<<"NG";

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.295497519686474x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

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

#define debug  freopen("in.txt","r",stdin),freopen("out.txt","w",stdout);

#define pb push_back

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

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int maxn = 1e6+10;

const int maxM = 1e6+10;

const int inf = 0x3f3f3f3f;



int K,A,B;

int main(){

	// debug;

	ios;



	cin>>K>>A>>B;

	if(B/K - (A-1)/K > 0) cout<<"OK\n";

	else cout<<"NG\n";



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.370092216486189x):
#include <stdio.h>



int main(){

	int a,b,c;	

	scanf("%d", &a);

	scanf("%d %d", &b, &c);

	

	for(int i = b; i<=c; i++){

		

		if (i%a == 0){

			printf("OK\n");

			return 0;

		}	

	}

	

	printf("NG\n");

	

return 0;

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

using namespace std;

int k,a,b,gg;

int main(){

	cin>>k>>a>>b;

	for (int i=1;i<=1000;i++){

		gg=k*i;

		if (a<=gg && gg<=b){

			cout<<"OK"<<endl;

			return 0;

		}

	}

	cout<<"NG"<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.293908162493575x):
#include<stdio.h>

int main(){

	 int K,A,B,i;

	 int Count= 0;

	 scanf("%d",&K);

	 scanf("%d %d",&A,&B);

	 for(i=A;i<=B;i++){

	 	if(i%K==0){

	 		Count++;

		 }

	 }	if(Count>0){

	 	     printf("OK\n");

	 }else{

	 	     printf("NG\n");

	 }

	 

	 

	 	

	 

	

	return 0;

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

using namespace std;

int k,a,b;

int main(){

	cin>>k>>a>>b;

	if(a/k<b/k || (a%k==0 || b%k==0)) cout<<"OK"<<endl;

	else cout<<"NG"<<endl;

	

} 

// Faster C++ solution for the same problem (measured speedup: 5.293943689189313x):
#include<stdio.h>

int main(){

	 int K,A,B,i;

	 int Count= 0;

	 scanf("%d",&K);

	 scanf("%d %d",&A,&B);

	 for(i=A;i<=B;i++){

	 	if(i%K==0){

	 		Count++;

		 }

	 }	if(Count>0){

	 	     printf("OK\n");

	 }else{

	 	     printf("NG\n");

	 }

	 

	 

	 	

	 

	

	return 0;

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

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

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

#define for0(i, n) for(int i = 0; i < n; ++i)

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

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

/*=================================================================================*/



int k, a, b, flag;



int main() {

   cin >> k >> a >> b;

   for(int i = a; i <= b; ++i) {

      if(!(i%k)) flag = 1;

   }

   flag ? puts("OK") : puts("NG");

}

// Faster C++ solution for the same problem (measured speedup: 5.22682599125647x):
#include<stdio.h>

int main(){

	 int K,A,B,i;

	 int Count= 0;

	 scanf("%d",&K);

	 scanf("%d %d",&A,&B);

	 for(i=A;i<=B;i++){

	 	if(i%K==0){

	 		Count++;

		 }

	 }	if(Count>0){

	 	     printf("OK\n");

	 }else{

	 	     printf("NG\n");

	 }

	 

	 

	 	

	 

	

	return 0;

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

using namespace std;

long long k,a,b;

int main(){

	cin>>k>>a>>b;

	if(b>=a/k*k+k||a==a/k*k||b==a/k*k+k) cout<<"OK"<<endl;

	else cout<<"NG"<<endl;

	return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 5.294062829075464x):
#include<stdio.h>

int main(){

	 int K,A,B,i;

	 int Count= 0;

	 scanf("%d",&K);

	 scanf("%d %d",&A,&B);

	 for(i=A;i<=B;i++){

	 	if(i%K==0){

	 		Count++;

		 }

	 }	if(Count>0){

	 	     printf("OK\n");

	 }else{

	 	     printf("NG\n");

	 }

	 

	 

	 	

	 

	

	return 0;

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

using namespace std;

int main()

{

    int k,a,b;

    cin>>k>>a>>b;

    for(int i=a;i<=b;i++){

        if(i%k==0){cout<<"OK";return 0;}

    }cout<<"NG";

}


// Faster C++ solution for the same problem (measured speedup: 5.280735442519425x):
#include<stdio.h>

int main(){

	 int K,A,B,i;

	 int Count= 0;

	 scanf("%d",&K);

	 scanf("%d %d",&A,&B);

	 for(i=A;i<=B;i++){

	 	if(i%K==0){

	 		Count++;

		 }

	 }	if(Count>0){

	 	     printf("OK\n");

	 }else{

	 	     printf("NG\n");

	 }

	 

	 

	 	

	 

	

	return 0;

}