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

using namespace std;

int main(){

    int k,a,b;

    cin>>k>>a>>b;

    for(int i=1;k*i<=b;i++){

        if(k*i>=a&&k*i<=b){

            cout<<"OK";

            return 0;

        }

    }

    cout<<"NG";

}


// Faster C++ solution for the same problem (measured speedup: 5.296814262325414x):
#include <stdio.h>



int main() {

    int abc;

    int a;

    int b;

    

    scanf("%d", &abc);

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

    while(a <= b) {

        if(a%abc == 0) {

            printf("OK");

            return 0;

        }

        a++;

    }

    printf("NG");

    

    return 0;

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

using namespace std;

int main(){

	int k,a,b;

	cin>>k>>a>>b;

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

		cout<<"OK"<<endl;

	}else{

		int x=b-a;

		int y=a%k;

		if(x+y>=k)cout<<"OK"<<endl;

		else cout<<"NG"<<endl;

	}

} 

// Faster C++ solution for the same problem (measured speedup: 5.310853799501425x):
#include <stdio.h>



int main() {

    int abc;

    int a;

    int b;

    

    scanf("%d", &abc);

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

    while(a <= b) {

        if(a%abc == 0) {

            printf("OK");

            return 0;

        }

        a++;

    }

    printf("NG");

    

    return 0;

}
// Slower C++ solution:
// /*

// power code taken from geeks for geeks

// */



#include <bits/stdc++.h>

#define MODD 1000000007LL

#define MAX_SIZE 100000000

#define MAX_N 200001

using namespace std;

typedef long long ll;

typedef long double ld;



ll max(ll a,ll b){return a>b?a:b;}

ll min(ll a,ll b){return a<b?a:b;}

double max(double a,double b){return a>b?a:b;}

double min(double a,double b){return a<b?a:b;}



ll modd(ll a){if(a<0LL) return a*-1LL;return a;}

double modd(double a){if(a<0.0) return a*-1.0;return a;}



ll modulo(ll a){return a % MODD; }



ll power(ll k, ll p){

    if (p == 0)

        return 1LL;

 

    ll P = power(k, p/2LL);

    P = (P * P) % MODD;

 

    if (p % 2 == 1)

        P = (P * k) % MODD;

 

    return P;

}





struct p_str{

    ll a,b,c;

};





void swap(ll& s1, ll &s2){

    ll temp=s1;

    s1=s2;

    s2=temp;

}

ll gcd(ll a, ll b){

    if(b==0){

        return a;

    }

    return gcd(b, a%b);

}



void solve(){

    ll k;

    cin>>k;

    ll a, b;

    cin>>a>>b;

    ll i=a;

    while(i<=b){

        if(i%k==0){

            cout<<"OK"<<endl;

            return;

        }

        i++;

    }

    cout<<"NG"<<endl;

}





int main(){





    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    

    ll t=1;



    // cin>>t;

    

    while(t--){

        solve();

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.376694633695773x):
#include <stdio.h>



int main() {

    int abc;

    int a;

    int b;

    

    scanf("%d", &abc);

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

    while(a <= b) {

        if(a%abc == 0) {

            printf("OK");

            return 0;

        }

        a++;

    }

    printf("NG");

    

    return 0;

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

using namespace std;



#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)

template <typename Arg1>

void __f(const char* name, Arg1&& arg1) {

  cerr << name << " : " << arg1 << "\n";

}

template <typename Arg1, typename... Args>

void __f(const char* na, Arg1&& arg1, Args&&... args) {

  const char* c = strchr(na + 1, ',');

  cerr.write(na, c - na) << " : " << arg1 << " , ";

  __f(c + 1, args...);

}

#define SZ(v) (int)v.size()

#define endl "\n"

#define ALL(v) v.begin(), v.end()

#define UNIQUE(v) sort(ALL(v)),v.erase(unique(ALL(v)),v.end())

typedef long long LL;

typedef pair <int,int> pii;

const double pi = acos(-1.0);

const double eps = 1e-8;

const int inf = 1e9;

const LL INF = 1e18;

const int mod = 1e9 + 7;

const int N = 1e5 + 5;



void solve() {

  int k, a, b;

  cin >> k >> a >> b;

  if (b/k - (a - 1)/k) cout << "OK" << endl;

  else cout << "NG" << endl;

}



int main() {

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

  #ifdef LOCAL

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

  #endif

  int t = 1;

  for (int tc = 1; tc <= t; tc++) solve();

  return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.3679791142247035x):
#include <stdio.h>



int main() {

    int abc;

    int a;

    int b;

    

    scanf("%d", &abc);

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

    while(a <= b) {

        if(a%abc == 0) {

            printf("OK");

            return 0;

        }

        a++;

    }

    printf("NG");

    

    return 0;

}
// Slower C++ solution:


#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,x,y,i=0,m=1;

    cin>>n>>x>>y;

    while(1)

    {

        i++;

        m=n*i;

        if(m>=x && m<=y)

            {

                cout<<"OK";

            return 0;

            }

        if(m>y)

            {

                cout<<"NG";

            return 0;

            }

    }



   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.296824415323517x):
#include <stdio.h>



int main() {

    int abc;

    int a;

    int b;

    

    scanf("%d", &abc);

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

    while(a <= b) {

        if(a%abc == 0) {

            printf("OK");

            return 0;

        }

        a++;

    }

    printf("NG");

    

    return 0;

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

using namespace std;

int main()

{

    int a,b,k,c=0;

    cin>>k>>a>>b;

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

    {

        if(i%k==0)

        {

            c=1;

            break;

        }

    }

    if(c==1)

        cout<<"OK\n";

    else

        cout<<"NG\n";

}


// Faster C++ solution for the same problem (measured speedup: 5.296833595921801x):
#include <stdio.h>



int main() {

    int abc;

    int a;

    int b;

    

    scanf("%d", &abc);

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

    while(a <= b) {

        if(a%abc == 0) {

            printf("OK");

            return 0;

        }

        a++;

    }

    printf("NG");

    

    return 0;

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

using namespace std;

int main()

{

  int a,b,k,x=0;

  cin>>k>>a>>b;

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

  {

      if(i%k==0){

        x=1;

        break;

        }

  }

if(x==1){

cout<<"OK";

}

else{

cout<<"NG";

}

}

// Faster C++ solution for the same problem (measured speedup: 5.296922127205255x):
#include <stdio.h>



int main() {

    int abc;

    int a;

    int b;

    

    scanf("%d", &abc);

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

    while(a <= b) {

        if(a%abc == 0) {

            printf("OK");

            return 0;

        }

        a++;

    }

    printf("NG");

    

    return 0;

}
// Slower C++ solution:
#pragma region 

#include<iostream>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<map>

#include<cstring>

#include<string>

typedef long long ll;

template <typename T>void read(T &x){x=0;char ch=getchar();ll f=1;while(!isdigit(ch)){if(ch=='-')f*=-1;ch=getchar();}while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}x*=f;}

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

using namespace std;

#pragma endregion



int main() {

    IO;

    int k, a, b;

    cin >> k >> a >> b;

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

        cout << "OK\n";

    else

        cout << (a / k - b / k ? "OK\n" : "NG\n");

}

// Faster C++ solution for the same problem (measured speedup: 5.371738887883862x):
#include <stdio.h>

#include <string.h>



int main(){

	int Tim, Min, Max;  //Times, Minimum, Maximum

	scanf("%d", &Tim);

	scanf("%d %d", &Min, &Max);

	

	for (Min==Min; Min<=Max; Min++){

		if (Min % Tim == 0){

			printf("OK\n");

			return 0;

		} else {

			continue;

		}



	}

	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=1;k*i<=b;i++){

        if(k*i>=a&&k*i<=b){

            cout<<"OK";

            return 0;

        }

    }

    cout<<"NG";

}


// Faster C++ solution for the same problem (measured speedup: 5.295008222196507x):
#include <stdio.h>

#include <string.h>



int main(){

	int Tim, Min, Max;  //Times, Minimum, Maximum

	scanf("%d", &Tim);

	scanf("%d %d", &Min, &Max);

	

	for (Min==Min; Min<=Max; Min++){

		if (Min % Tim == 0){

			printf("OK\n");

			return 0;

		} else {

			continue;

		}



	}

	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;

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

		cout<<"OK"<<endl;

	}else{

		int x=b-a;

		int y=a%k;

		if(x+y>=k)cout<<"OK"<<endl;

		else cout<<"NG"<<endl;

	}

} 

// Faster C++ solution for the same problem (measured speedup: 5.309042972350304x):
#include <stdio.h>

#include <string.h>



int main(){

	int Tim, Min, Max;  //Times, Minimum, Maximum

	scanf("%d", &Tim);

	scanf("%d %d", &Min, &Max);

	

	for (Min==Min; Min<=Max; Min++){

		if (Min % Tim == 0){

			printf("OK\n");

			return 0;

		} else {

			continue;

		}



	}

	printf ("NG\n");

return 0;

}
// Slower C++ solution:
// /*

// power code taken from geeks for geeks

// */



#include <bits/stdc++.h>

#define MODD 1000000007LL

#define MAX_SIZE 100000000

#define MAX_N 200001

using namespace std;

typedef long long ll;

typedef long double ld;



ll max(ll a,ll b){return a>b?a:b;}

ll min(ll a,ll b){return a<b?a:b;}

double max(double a,double b){return a>b?a:b;}

double min(double a,double b){return a<b?a:b;}



ll modd(ll a){if(a<0LL) return a*-1LL;return a;}

double modd(double a){if(a<0.0) return a*-1.0;return a;}



ll modulo(ll a){return a % MODD; }



ll power(ll k, ll p){

    if (p == 0)

        return 1LL;

 

    ll P = power(k, p/2LL);

    P = (P * P) % MODD;

 

    if (p % 2 == 1)

        P = (P * k) % MODD;

 

    return P;

}





struct p_str{

    ll a,b,c;

};





void swap(ll& s1, ll &s2){

    ll temp=s1;

    s1=s2;

    s2=temp;

}

ll gcd(ll a, ll b){

    if(b==0){

        return a;

    }

    return gcd(b, a%b);

}



void solve(){

    ll k;

    cin>>k;

    ll a, b;

    cin>>a>>b;

    ll i=a;

    while(i<=b){

        if(i%k==0){

            cout<<"OK"<<endl;

            return;

        }

        i++;

    }

    cout<<"NG"<<endl;

}





int main(){





    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    

    ll t=1;



    // cin>>t;

    

    while(t--){

        solve();

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.374861356977274x):
#include <stdio.h>

#include <string.h>



int main(){

	int Tim, Min, Max;  //Times, Minimum, Maximum

	scanf("%d", &Tim);

	scanf("%d %d", &Min, &Max);

	

	for (Min==Min; Min<=Max; Min++){

		if (Min % Tim == 0){

			printf("OK\n");

			return 0;

		} else {

			continue;

		}



	}

	printf ("NG\n");

return 0;

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

using namespace std;



#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)

template <typename Arg1>

void __f(const char* name, Arg1&& arg1) {

  cerr << name << " : " << arg1 << "\n";

}

template <typename Arg1, typename... Args>

void __f(const char* na, Arg1&& arg1, Args&&... args) {

  const char* c = strchr(na + 1, ',');

  cerr.write(na, c - na) << " : " << arg1 << " , ";

  __f(c + 1, args...);

}

#define SZ(v) (int)v.size()

#define endl "\n"

#define ALL(v) v.begin(), v.end()

#define UNIQUE(v) sort(ALL(v)),v.erase(unique(ALL(v)),v.end())

typedef long long LL;

typedef pair <int,int> pii;

const double pi = acos(-1.0);

const double eps = 1e-8;

const int inf = 1e9;

const LL INF = 1e18;

const int mod = 1e9 + 7;

const int N = 1e5 + 5;



void solve() {

  int k, a, b;

  cin >> k >> a >> b;

  if (b/k - (a - 1)/k) cout << "OK" << endl;

  else cout << "NG" << endl;

}



int main() {

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

  #ifdef LOCAL

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

  #endif

  int t = 1;

  for (int tc = 1; tc <= t; tc++) solve();

  return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.366148809212805x):
#include <stdio.h>

#include <string.h>



int main(){

	int Tim, Min, Max;  //Times, Minimum, Maximum

	scanf("%d", &Tim);

	scanf("%d %d", &Min, &Max);

	

	for (Min==Min; Min<=Max; Min++){

		if (Min % Tim == 0){

			printf("OK\n");

			return 0;

		} else {

			continue;

		}



	}

	printf ("NG\n");

return 0;

}
// Slower C++ solution:


#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,x,y,i=0,m=1;

    cin>>n>>x>>y;

    while(1)

    {

        i++;

        m=n*i;

        if(m>=x && m<=y)

            {

                cout<<"OK";

            return 0;

            }

        if(m>y)

            {

                cout<<"NG";

            return 0;

            }

    }



   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.29501837173277x):
#include <stdio.h>

#include <string.h>



int main(){

	int Tim, Min, Max;  //Times, Minimum, Maximum

	scanf("%d", &Tim);

	scanf("%d %d", &Min, &Max);

	

	for (Min==Min; Min<=Max; Min++){

		if (Min % Tim == 0){

			printf("OK\n");

			return 0;

		} else {

			continue;

		}



	}

	printf ("NG\n");

return 0;

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

using namespace std;

int main()

{

    int a,b,k,c=0;

    cin>>k>>a>>b;

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

    {

        if(i%k==0)

        {

            c=1;

            break;

        }

    }

    if(c==1)

        cout<<"OK\n";

    else

        cout<<"NG\n";

}


// Faster C++ solution for the same problem (measured speedup: 5.29502754920077x):
#include <stdio.h>

#include <string.h>



int main(){

	int Tim, Min, Max;  //Times, Minimum, Maximum

	scanf("%d", &Tim);

	scanf("%d %d", &Min, &Max);

	

	for (Min==Min; Min<=Max; Min++){

		if (Min % Tim == 0){

			printf("OK\n");

			return 0;

		} else {

			continue;

		}



	}

	printf ("NG\n");

return 0;

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

using namespace std;

int main()

{

  int a,b,k,x=0;

  cin>>k>>a>>b;

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

  {

      if(i%k==0){

        x=1;

        break;

        }

  }

if(x==1){

cout<<"OK";

}

else{

cout<<"NG";

}

}

// Faster C++ solution for the same problem (measured speedup: 5.295116050297957x):
#include <stdio.h>

#include <string.h>



int main(){

	int Tim, Min, Max;  //Times, Minimum, Maximum

	scanf("%d", &Tim);

	scanf("%d %d", &Min, &Max);

	

	for (Min==Min; Min<=Max; Min++){

		if (Min % Tim == 0){

			printf("OK\n");

			return 0;

		} else {

			continue;

		}



	}

	printf ("NG\n");

return 0;

}
// Slower C++ solution:
#pragma region 

#include<iostream>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<map>

#include<cstring>

#include<string>

typedef long long ll;

template <typename T>void read(T &x){x=0;char ch=getchar();ll f=1;while(!isdigit(ch)){if(ch=='-')f*=-1;ch=getchar();}while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}x*=f;}

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

using namespace std;

#pragma endregion



int main() {

    IO;

    int k, a, b;

    cin >> k >> a >> b;

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

        cout << "OK\n";

    else

        cout << (a / k - b / k ? "OK\n" : "NG\n");

}

// Faster C++ solution for the same problem (measured speedup: 5.357029519068189x):
#include <stdio.h>



int main(){

	int amount, first, second;

	int mark = 0;

	

	scanf("%d", &amount);

	scanf("%d %d", &first, &second);

	

	for(int i = first; i <= second; i++){

		if(i % amount == 0){

			mark = 1;

		}

	}

	

	if(mark == 1){

		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=1;k*i<=b;i++){

        if(k*i>=a&&k*i<=b){

            cout<<"OK";

            return 0;

        }

    }

    cout<<"NG";

}


// Faster C++ solution for the same problem (measured speedup: 5.280508964051629x):
#include <stdio.h>



int main(){

	int amount, first, second;

	int mark = 0;

	

	scanf("%d", &amount);

	scanf("%d %d", &first, &second);

	

	for(int i = first; i <= second; i++){

		if(i % amount == 0){

			mark = 1;

		}

	}

	

	if(mark == 1){

		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;

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

		cout<<"OK"<<endl;

	}else{

		int x=b-a;

		int y=a%k;

		if(x+y>=k)cout<<"OK"<<endl;

		else cout<<"NG"<<endl;

	}

} 

// Faster C++ solution for the same problem (measured speedup: 5.294505283015721x):
#include <stdio.h>



int main(){

	int amount, first, second;

	int mark = 0;

	

	scanf("%d", &amount);

	scanf("%d %d", &first, &second);

	

	for(int i = first; i <= second; i++){

		if(i % amount == 0){

			mark = 1;

		}

	}

	

	if(mark == 1){

		printf("OK\n");

	}else{

		printf("NG\n");

	}

	return 0;

}
// Slower C++ solution:
// /*

// power code taken from geeks for geeks

// */



#include <bits/stdc++.h>

#define MODD 1000000007LL

#define MAX_SIZE 100000000

#define MAX_N 200001

using namespace std;

typedef long long ll;

typedef long double ld;



ll max(ll a,ll b){return a>b?a:b;}

ll min(ll a,ll b){return a<b?a:b;}

double max(double a,double b){return a>b?a:b;}

double min(double a,double b){return a<b?a:b;}



ll modd(ll a){if(a<0LL) return a*-1LL;return a;}

double modd(double a){if(a<0.0) return a*-1.0;return a;}



ll modulo(ll a){return a % MODD; }



ll power(ll k, ll p){

    if (p == 0)

        return 1LL;

 

    ll P = power(k, p/2LL);

    P = (P * P) % MODD;

 

    if (p % 2 == 1)

        P = (P * k) % MODD;

 

    return P;

}





struct p_str{

    ll a,b,c;

};





void swap(ll& s1, ll &s2){

    ll temp=s1;

    s1=s2;

    s2=temp;

}

ll gcd(ll a, ll b){

    if(b==0){

        return a;

    }

    return gcd(b, a%b);

}



void solve(){

    ll k;

    cin>>k;

    ll a, b;

    cin>>a>>b;

    ll i=a;

    while(i<=b){

        if(i%k==0){

            cout<<"OK"<<endl;

            return;

        }

        i++;

    }

    cout<<"NG"<<endl;

}





int main(){





    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    

    ll t=1;



    // cin>>t;

    

    while(t--){

        solve();

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.360143437941557x):
#include <stdio.h>



int main(){

	int amount, first, second;

	int mark = 0;

	

	scanf("%d", &amount);

	scanf("%d %d", &first, &second);

	

	for(int i = first; i <= second; i++){

		if(i % amount == 0){

			mark = 1;

		}

	}

	

	if(mark == 1){

		printf("OK\n");

	}else{

		printf("NG\n");

	}

	return 0;

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

using namespace std;



#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)

template <typename Arg1>

void __f(const char* name, Arg1&& arg1) {

  cerr << name << " : " << arg1 << "\n";

}

template <typename Arg1, typename... Args>

void __f(const char* na, Arg1&& arg1, Args&&... args) {

  const char* c = strchr(na + 1, ',');

  cerr.write(na, c - na) << " : " << arg1 << " , ";

  __f(c + 1, args...);

}

#define SZ(v) (int)v.size()

#define endl "\n"

#define ALL(v) v.begin(), v.end()

#define UNIQUE(v) sort(ALL(v)),v.erase(unique(ALL(v)),v.end())

typedef long long LL;

typedef pair <int,int> pii;

const double pi = acos(-1.0);

const double eps = 1e-8;

const int inf = 1e9;

const LL INF = 1e18;

const int mod = 1e9 + 7;

const int N = 1e5 + 5;



void solve() {

  int k, a, b;

  cin >> k >> a >> b;

  if (b/k - (a - 1)/k) cout << "OK" << endl;

  else cout << "NG" << endl;

}



int main() {

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

  #ifdef LOCAL

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

  #endif

  int t = 1;

  for (int tc = 1; tc <= t; tc++) solve();

  return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.351454747643184x):
#include <stdio.h>



int main(){

	int amount, first, second;

	int mark = 0;

	

	scanf("%d", &amount);

	scanf("%d %d", &first, &second);

	

	for(int i = first; i <= second; i++){

		if(i % amount == 0){

			mark = 1;

		}

	}

	

	if(mark == 1){

		printf("OK\n");

	}else{

		printf("NG\n");

	}

	return 0;

}
// Slower C++ solution:


#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,x,y,i=0,m=1;

    cin>>n>>x>>y;

    while(1)

    {

        i++;

        m=n*i;

        if(m>=x && m<=y)

            {

                cout<<"OK";

            return 0;

            }

        if(m>y)

            {

                cout<<"NG";

            return 0;

            }

    }



   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.280519085795537x):
#include <stdio.h>



int main(){

	int amount, first, second;

	int mark = 0;

	

	scanf("%d", &amount);

	scanf("%d %d", &first, &second);

	

	for(int i = first; i <= second; i++){

		if(i % amount == 0){

			mark = 1;

		}

	}

	

	if(mark == 1){

		printf("OK\n");

	}else{

		printf("NG\n");

	}

	return 0;

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

using namespace std;

int main()

{

    int a,b,k,c=0;

    cin>>k>>a>>b;

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

    {

        if(i%k==0)

        {

            c=1;

            break;

        }

    }

    if(c==1)

        cout<<"OK\n";

    else

        cout<<"NG\n";

}


// Faster C++ solution for the same problem (measured speedup: 5.280528238132986x):
#include <stdio.h>



int main(){

	int amount, first, second;

	int mark = 0;

	

	scanf("%d", &amount);

	scanf("%d %d", &first, &second);

	

	for(int i = first; i <= second; i++){

		if(i % amount == 0){

			mark = 1;

		}

	}

	

	if(mark == 1){

		printf("OK\n");

	}else{

		printf("NG\n");

	}

	return 0;

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

using namespace std;

int main()

{

  int a,b,k,x=0;

  cin>>k>>a>>b;

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

  {

      if(i%k==0){

        x=1;

        break;

        }

  }

if(x==1){

cout<<"OK";

}

else{

cout<<"NG";

}

}

// Faster C++ solution for the same problem (measured speedup: 5.280616496888668x):
#include <stdio.h>



int main(){

	int amount, first, second;

	int mark = 0;

	

	scanf("%d", &amount);

	scanf("%d %d", &first, &second);

	

	for(int i = first; i <= second; i++){

		if(i % amount == 0){

			mark = 1;

		}

	}

	

	if(mark == 1){

		printf("OK\n");

	}else{

		printf("NG\n");

	}

	return 0;

}
// Slower C++ solution:
#pragma region 

#include<iostream>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<map>

#include<cstring>

#include<string>

typedef long long ll;

template <typename T>void read(T &x){x=0;char ch=getchar();ll f=1;while(!isdigit(ch)){if(ch=='-')f*=-1;ch=getchar();}while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}x*=f;}

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

using namespace std;

#pragma endregion



int main() {

    IO;

    int k, a, b;

    cin >> k >> a >> b;

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

        cout << "OK\n";

    else

        cout << (a / k - b / k ? "OK\n" : "NG\n");

}

// Faster C++ solution for the same problem (measured speedup: 5.376155848918361x):
#include<stdio.h>



int main(){

	int kelipatan,batas1,batas2;

	scanf("%d %d %d", &kelipatan,&batas1,&batas2);

	int temp=batas2/kelipatan;

	if(temp*kelipatan>=batas1){

		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=1;k*i<=b;i++){

        if(k*i>=a&&k*i<=b){

            cout<<"OK";

            return 0;

        }

    }

    cout<<"NG";

}


// Faster C++ solution for the same problem (measured speedup: 5.2993620907450225x):
#include<stdio.h>



int main(){

	int kelipatan,batas1,batas2;

	scanf("%d %d %d", &kelipatan,&batas1,&batas2);

	int temp=batas2/kelipatan;

	if(temp*kelipatan>=batas1){

		printf("OK\n");

	}

	else printf("NG\n");

	return 0;

}