stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 5.308834290362716x):
#include <stdio.h>

int main (){

	int n;

	scanf ("%d", &n);

	int c, r;

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

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

		if (i % n == 0){

			printf ("OK");

			return 0;

		}else{

		continue;

		}

	}

	printf ("NG");

	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: 5.295469717815326x):
#include <stdio.h>

int main (){

	int n;

	scanf ("%d", &n);

	int c, r;

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

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

		if (i % n == 0){

			printf ("OK");

			return 0;

		}else{

		continue;

		}

	}

	printf ("NG");

	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: 5.29550961816424x):
#include <stdio.h>

int main (){

	int n;

	scanf ("%d", &n);

	int c, r;

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

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

		if (i % n == 0){

			printf ("OK");

			return 0;

		}else{

		continue;

		}

	}

	printf ("NG");

	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: 5.309567321674282x):
#include <stdio.h>

int main (){

	int n;

	scanf ("%d", &n);

	int c, r;

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

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

		if (i % n == 0){

			printf ("OK");

			return 0;

		}else{

		continue;

		}

	}

	printf ("NG");

	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: 5.309646107350695x):
#include <stdio.h>

int main (){

	int n;

	scanf ("%d", &n);

	int c, r;

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

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

		if (i % n == 0){

			printf ("OK");

			return 0;

		}else{

		continue;

		}

	}

	printf ("NG");

	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: 5.294272464314599x):
#include <stdio.h>

int main (){

	int n;

	scanf ("%d", &n);

	int c, r;

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

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

		if (i % n == 0){

			printf ("OK");

			return 0;

		}else{

		continue;

		}

	}

	printf ("NG");

	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: 5.310792624076683x):
#include <stdio.h>

int main (){

	int n;

	scanf ("%d", &n);

	int c, r;

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

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

		if (i % n == 0){

			printf ("OK");

			return 0;

		}else{

		continue;

		}

	}

	printf ("NG");

	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: 5.295509632460316x):
#include <stdio.h>

int main (){

	int n;

	scanf ("%d", &n);

	int c, r;

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

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

		if (i % n == 0){

			printf ("OK");

			return 0;

		}else{

		continue;

		}

	}

	printf ("NG");

	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: 5.370104499885867x):
#include <stdio.h>

int main (){

	int n;

	scanf ("%d", &n);

	int c, r;

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

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

		if (i % n == 0){

			printf ("OK");

			return 0;

		}else{

		continue;

		}

	}

	printf ("NG");

	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: 5.310542646736166x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		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: 5.310578285063659x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		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: 5.24324969033125x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		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: 5.310697799310396x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		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: 5.297328535525878x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		printf("OK\n");

	}	else {

		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: 5.2973684498806275x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		printf("OK\n");

	}	else {

		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: 5.31143108793088x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		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,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: 5.311509901262669x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		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,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: 5.296130861764791x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		printf("OK\n");

	}	else {

		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: 5.312656820439376x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		printf("OK\n");

	}	else {

		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: 5.297368464181722x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		printf("OK\n");

	}	else {

		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: 5.3719895159248265x):
#include <stdio.h>

//jarak ke titik adalah kelipatan k

//a antara b

//si antara itu hrs terpenuhi

//kl terpenuhi ok, klk ga ng

//k = kelipatan

//a titik awal, b ujung

//7 

//500 600

//600/7 = 85

//85*7 = 595

//baru dicek

//4

// 5 7

//7/4 = 1

//1 *4 = 5

//kalo lebih kecil dari bates 1 jadinya false

int main(){

	int kelipatan, awal, akhir, patokan;

	

	scanf("%d", &kelipatan);

	scanf("%d %d", &awal, &akhir);

	

	patokan = akhir/kelipatan;

	

	if(patokan*kelipatan >= awal){

		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: 5.295701067503803x):
#include <stdio.h>

int main(){

    int K, A, B;

    scanf("%d", &K);

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

    int L = B - A;

    int count = 0;

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

        int total = A % K;

        A++;

        if(total == 0){

            count++;

        }

        else{

            count = count + 0;

    	}

	}

    

	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: 5.295736606231482x):
#include <stdio.h>

int main(){

    int K, A, B;

    scanf("%d", &K);

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

    int L = B - A;

    int count = 0;

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

        int total = A % K;

        A++;

        if(total == 0){

            count++;

        }

        else{

            count = count + 0;

    	}

	}

    

	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: 5.228596177330664x):
#include <stdio.h>

int main(){

    int K, A, B;

    scanf("%d", &K);

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

    int L = B - A;

    int count = 0;

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

        int total = A % K;

        A++;

        if(total == 0){

            count++;

        }

        else{

            count = count + 0;

    	}

	}

    

	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: 5.295855786467124x):
#include <stdio.h>

int main(){

    int K, A, B;

    scanf("%d", &K);

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

    int L = B - A;

    int count = 0;

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

        int total = A % K;

        A++;

        if(total == 0){

            count++;

        }

        else{

            count = count + 0;

    	}

	}

    

	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: 5.282523886281947x):
#include <stdio.h>

int main(){

    int K, A, B;

    scanf("%d", &K);

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

    int L = B - A;

    int count = 0;

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

        int total = A % K;

        A++;

        if(total == 0){

            count++;

        }

        else{

            count = count + 0;

    	}

	}

    

	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 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: 5.282563689086503x):
#include <stdio.h>

int main(){

    int K, A, B;

    scanf("%d", &K);

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

    int L = B - A;

    int count = 0;

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

        int total = A % K;

        A++;

        if(total == 0){

            count++;

        }

        else{

            count = count + 0;

    	}

	}

    

	if(count > 0){

        printf("OK\n");

    }

    else{

        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: 5.296587025737497x):
#include <stdio.h>

int main(){

    int K, A, B;

    scanf("%d", &K);

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

    int L = B - A;

    int count = 0;

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

        int total = A % K;

        A++;

        if(total == 0){

            count++;

        }

        else{

            count = count + 0;

    	}

	}

    

	if(count > 0){

        printf("OK\n");

    }

    else{

        printf("NG\n");

    }

	

 return 0;

}