#include <bits/stdc++.h>
#define XOX
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std;
#define Time  cerr << "Time Taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl
#define ll  long long
#define beg() begin()

#define test int t=1; cin>>t;  while(t--)
#define vll vector<long long>
#define vi vector<int>
#define all(v) v.begin(),v.end()
#define pb push_back
#define forr(i,a,b) for (ll i = (a); i < (b); ++i)
#define rrof(i,a,b) for (ll i = (b)-1; i >= (a); --i)

#define mod 1000000007 
void solve( );
void swap(ll &a, ll &b);
ll min(ll &a, ll &b);
ll max(ll &a, ll &b);
void exp(ll b);
ll sqrt(ll a);
ll  GCD(ll a, ll b) ;


int main() {

	fast;
	#ifndef ONLINE_JUDGE
   		freopen("input.txt", "r", stdin);
   		freopen("output.txt", "w", stdout);
	#endif
   
	test{ solve(); }
   		//solve();
Time;
 
  return 0;
} 
 
void solve(){

	ll n,k; cin>>n>>k; 
	if(n>k){ 

		if(n%k != 0) cout<< 2 <<"\n";
		else cout<< 1 <<"\n";
	}else{
		if(k%n != 0) cout<<k/n + 1<<"\n";
		else cout<< k/n <<"\n";
	}

 
 }
void swap(ll &a, ll &b){ 
	if(a==b) return ;

	a=a^b;
	b=a^b;
	a=a^b;
	return ;
}

ll min(ll &a, ll &b){
	if(a>b) return b;
	return a;
}
ll max(ll &a, ll &b){
	if(a<b) return b;
	return a;
}

ll sqrt(ll a){
	ll l=1,h=a, m=l+(h-l+1)/2;
	while(l<h){
		if(m > a/m ) h=m-1;
		else l=m;

		m=l+(h-l+1)/2;
	}

	return m; 
} 
ll  GCD(ll a, ll b) {
  if(b==0) return a;
  else return GCD(b, a%b) ; 
}


### Prompt

Please formulate a Cpp solution 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
#include <bits/stdc++.h>
#define XOX
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std;
#define Time  cerr << "Time Taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl
#define ll  long long
#define beg() begin()

#define test int t=1; cin>>t;  while(t--)
#define vll vector<long long>
#define vi vector<int>
#define all(v) v.begin(),v.end()
#define pb push_back
#define forr(i,a,b) for (ll i = (a); i < (b); ++i)
#define rrof(i,a,b) for (ll i = (b)-1; i >= (a); --i)

#define mod 1000000007 
void solve( );
void swap(ll &a, ll &b);
ll min(ll &a, ll &b);
ll max(ll &a, ll &b);
void exp(ll b);
ll sqrt(ll a);
ll  GCD(ll a, ll b) ;


int main() {

	fast;
	#ifndef ONLINE_JUDGE
   		freopen("input.txt", "r", stdin);
   		freopen("output.txt", "w", stdout);
	#endif
   
	test{ solve(); }
   		//solve();
Time;
 
  return 0;
} 
 
void solve(){

	ll n,k; cin>>n>>k; 
	if(n>k){ 

		if(n%k != 0) cout<< 2 <<"\n";
		else cout<< 1 <<"\n";
	}else{
		if(k%n != 0) cout<<k/n + 1<<"\n";
		else cout<< k/n <<"\n";
	}

 
 }
void swap(ll &a, ll &b){ 
	if(a==b) return ;

	a=a^b;
	b=a^b;
	a=a^b;
	return ;
}

ll min(ll &a, ll &b){
	if(a>b) return b;
	return a;
}
ll max(ll &a, ll &b){
	if(a<b) return b;
	return a;
}

ll sqrt(ll a){
	ll l=1,h=a, m=l+(h-l+1)/2;
	while(l<h){
		if(m > a/m ) h=m-1;
		else l=m;

		m=l+(h-l+1)/2;
	}

	return m; 
} 
ll  GCD(ll a, ll b) {
  if(b==0) return a;
  else return GCD(b, a%b) ; 
}


```