// Coded by register-Gen
// Date: 2021-01-29
// Special thanks to cf tool by xalanq

#include<bits/stdc++.h>

#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
#define Grep(u) for(int i=head[u];i;i=e[i].nxt)

#define clear(a) memset(a,0,sizeof(a))
#define setinf(a) memset(a,0x3f,sizeof(a))
#define setninf(a) memset(a,0xcf,sizeof(a))
#define setneg1(a) memset(a,-1,sizeof(a))

#define pb push_back
#define mp std::make_pair
#define fi first
#define se second

#define yes {puts("YES");return;}
#define no {puts("NO");return;}

typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef std::pair<int,int> pii;
typedef std::pair<ll,ll> pll;
typedef std::vector<int> vi;
typedef std::vector<ll> vl;

const int N=5e5;
const int M=1e6;
const int P=998244353;
const int inf=0x3f3f3f3f;
const int ninf=0xcfcfcfcf;
const ll llinf=0x3f3f3f3f3f3f3f3f;
const ll llninf=0xcfcfcfcfcfcfcfcf;

struct Edge{int to,nxt;}e[M*2+10];int head[N+10],tote;
inline void addEdge(int u,int v){e[++tote].to=v;e[tote].nxt=head[u];head[u]=tote;}

//struct Edge{int to,nxt,w;}e[M*2+10];int head[N+10],tote;
//inline void addEdge(int u,int v,int w){e[++tote].to=v;e[tote].w=w;e[tote].nxt=head[u];head[u]=tote;}

#define gc() char(getchar())

template<typename T>inline T read(){
	T x=0;bool f=0;char c=gc();
	while(c<'0'||c>'9')f|=c=='-',c=gc();
	while(c>='0'&&c<='9')x=x*10+c-48,c=gc();
	return f?-x:x;
}
inline int ri(){return read<int>();}
inline ll rl(){return read<ll>();}
inline ui ru(){return read<ui>();}
inline ull rul(){return read<ull>();}
inline int rstr(char*& str){
	char c=gc();int res=0;
	while(c!='\n'&&c!='\r'&&c!=' ')str[++res]=c,c=gc();
	return res;
}
inline int rline(char*& str){
	char c=gc();int res=0;
	while(c!='\n'&&c!='\r')str[++res]=c,c=gc();
	return res;
}

#undef gc

template<typename T1,typename T2>inline void chkmax(T1& a,T2 b){if(a<b)a=b;}
template<typename T1,typename T2>inline void chkmin(T1& a,T2 b){if(a>b)a=b;}
template<typename T1,typename T2,typename T3>inline void chkmax(T1& a,T2& b,T3 c){if(c>a){b=a;a=c;}else if(c>b)b=c;}
template<typename T1,typename T2,typename T3>inline void chkmin(T1& a,T2& b,T3 c){if(c<a){b=a;a=c;}else if(c<b)b=c;}
template<typename T1,typename T2,typename T3>inline void add(T1& x,T2 y,T3 p=P){(x+=y)>=p&&(x-=p);}
template<typename T1,typename T2,typename T3>inline void mul(T1& x,T2 y,T3 p=P){x=1LL*x*y%P;}

template<typename T1,typename T2,typename T3>inline T1 qpow(T1 x,T2 y,T3 p=P){
	T1 res=T1(1);
	while(y){if(y&1)mul(res,x,p);y>>=1;mul(x,x,p);}
	return res;
}
template<typename T1,typename T2,typename T3>inline T1 qmul(T1 x,T2 y,T3 p=P){
	T1 res=T1(0);
	while(y){if(y&1)add(res,x,p);y>>=1;add(x,x,p);}
	return res;
}
template<typename T1,typename T2>inline T1 gcd(T1 x,T2 y){return !y?x:gcd(y,T2(x%y));}
template<typename T1,typename T2,typename T3>inline T1 exgcd(T1 a,T2 b,T3& x,T3& y){
	if(b==0)return x=T3(1),y=T3(0),void();
	T1 res=exgcd(b,a%b,x,y);
	T3 tmp=x;x=y;y=tmp-(a/b)*y;
	return res;
}

void solve();
int main(){
	int T;
//	T=1;
	T=ri();
	while(T--)solve();
	return 0;
}

//
ll n,k;

void solve(){
	n=rl(),k=rl();
	if(n<=k)printf("%lld\n",ll(ceil(1.0*k/n)));
	else{
		ll t=ll(ceil(1.0*n/k));
		k=k*t;
		printf("%lld\n",ll(ceil(1.0*k/n)));
	}
}
### Prompt

Please create a solution in CPP to the following problem:

You are given two integers n and k.

You should create an array of n positive integers a_1, a_2, ..., a_n such that the sum (a_1 + a_2 + ... + a_n) is divisible by k and maximum element in a is minimum possible.

What is the minimum possible maximum element in a?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first and only line of each test case contains two integers n and k (1 ≤ n ≤ 10^9; 1 ≤ k ≤ 10^9).

Output

For each test case, print one integer — the minimum possible maximum element in array a such that the sum (a_1 + ... + a_n) is divisible by k. 

Example

Input


4
1 5
4 3
8 8
8 17


Output


5
2
1
3

Note

In the first test case n = 1, so the array consists of one element a_1 and if we make a_1 = 5 it will be divisible by k = 5 and the minimum possible.

In the second test case, we can create array a = [1, 2, 1, 2]. The sum is divisible by k = 3 and the maximum is equal to 2.

In the third test case, we can create array a = [1, 1, 1, 1, 1, 1, 1, 1]. The sum is divisible by k = 8 and the maximum is equal to 1.

### Response

```cpp
// Coded by register-Gen
// Date: 2021-01-29
// Special thanks to cf tool by xalanq

#include<bits/stdc++.h>

#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
#define Grep(u) for(int i=head[u];i;i=e[i].nxt)

#define clear(a) memset(a,0,sizeof(a))
#define setinf(a) memset(a,0x3f,sizeof(a))
#define setninf(a) memset(a,0xcf,sizeof(a))
#define setneg1(a) memset(a,-1,sizeof(a))

#define pb push_back
#define mp std::make_pair
#define fi first
#define se second

#define yes {puts("YES");return;}
#define no {puts("NO");return;}

typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef std::pair<int,int> pii;
typedef std::pair<ll,ll> pll;
typedef std::vector<int> vi;
typedef std::vector<ll> vl;

const int N=5e5;
const int M=1e6;
const int P=998244353;
const int inf=0x3f3f3f3f;
const int ninf=0xcfcfcfcf;
const ll llinf=0x3f3f3f3f3f3f3f3f;
const ll llninf=0xcfcfcfcfcfcfcfcf;

struct Edge{int to,nxt;}e[M*2+10];int head[N+10],tote;
inline void addEdge(int u,int v){e[++tote].to=v;e[tote].nxt=head[u];head[u]=tote;}

//struct Edge{int to,nxt,w;}e[M*2+10];int head[N+10],tote;
//inline void addEdge(int u,int v,int w){e[++tote].to=v;e[tote].w=w;e[tote].nxt=head[u];head[u]=tote;}

#define gc() char(getchar())

template<typename T>inline T read(){
	T x=0;bool f=0;char c=gc();
	while(c<'0'||c>'9')f|=c=='-',c=gc();
	while(c>='0'&&c<='9')x=x*10+c-48,c=gc();
	return f?-x:x;
}
inline int ri(){return read<int>();}
inline ll rl(){return read<ll>();}
inline ui ru(){return read<ui>();}
inline ull rul(){return read<ull>();}
inline int rstr(char*& str){
	char c=gc();int res=0;
	while(c!='\n'&&c!='\r'&&c!=' ')str[++res]=c,c=gc();
	return res;
}
inline int rline(char*& str){
	char c=gc();int res=0;
	while(c!='\n'&&c!='\r')str[++res]=c,c=gc();
	return res;
}

#undef gc

template<typename T1,typename T2>inline void chkmax(T1& a,T2 b){if(a<b)a=b;}
template<typename T1,typename T2>inline void chkmin(T1& a,T2 b){if(a>b)a=b;}
template<typename T1,typename T2,typename T3>inline void chkmax(T1& a,T2& b,T3 c){if(c>a){b=a;a=c;}else if(c>b)b=c;}
template<typename T1,typename T2,typename T3>inline void chkmin(T1& a,T2& b,T3 c){if(c<a){b=a;a=c;}else if(c<b)b=c;}
template<typename T1,typename T2,typename T3>inline void add(T1& x,T2 y,T3 p=P){(x+=y)>=p&&(x-=p);}
template<typename T1,typename T2,typename T3>inline void mul(T1& x,T2 y,T3 p=P){x=1LL*x*y%P;}

template<typename T1,typename T2,typename T3>inline T1 qpow(T1 x,T2 y,T3 p=P){
	T1 res=T1(1);
	while(y){if(y&1)mul(res,x,p);y>>=1;mul(x,x,p);}
	return res;
}
template<typename T1,typename T2,typename T3>inline T1 qmul(T1 x,T2 y,T3 p=P){
	T1 res=T1(0);
	while(y){if(y&1)add(res,x,p);y>>=1;add(x,x,p);}
	return res;
}
template<typename T1,typename T2>inline T1 gcd(T1 x,T2 y){return !y?x:gcd(y,T2(x%y));}
template<typename T1,typename T2,typename T3>inline T1 exgcd(T1 a,T2 b,T3& x,T3& y){
	if(b==0)return x=T3(1),y=T3(0),void();
	T1 res=exgcd(b,a%b,x,y);
	T3 tmp=x;x=y;y=tmp-(a/b)*y;
	return res;
}

void solve();
int main(){
	int T;
//	T=1;
	T=ri();
	while(T--)solve();
	return 0;
}

//
ll n,k;

void solve(){
	n=rl(),k=rl();
	if(n<=k)printf("%lld\n",ll(ceil(1.0*k/n)));
	else{
		ll t=ll(ceil(1.0*n/k));
		k=k*t;
		printf("%lld\n",ll(ceil(1.0*k/n)));
	}
}
```