////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
#include<bits/stdc++.h>
#define PIE acos(-1)
#define sa(arr,n)    cout<<"-------arr---------\n";rl(i,n)cout<<i<<" "<<arr[i]<<"\n";cout<<"\n";
#define sac(arr,j,n) cout<<"-------arr--------\n";rlc(i,j,n)cout<<i<<" "<<arr[i]<<"\n";cout<<"\n";
#define rlc(i,a,n)     for(int i=a;i<n;i++)   //run_loop
#define rl(i,n)        for(int i=0;i<n;i++)  //run_loop2
#define farr(arr,n)   rl(i,n)cin>>arr[i]   //fill_array
#define farrc(arr,a,n) rlc(i,a,n)cin>>arr[i]
#define ll             long long int
#define ld             long double
#define pb             push_back
#define all(v)         v.begin(),v.end()
#define eb             emplace_back
#define memas(a)        memset(a,-1,sizeof(a))   //memset_assign.
#define mema0(a)        memset(a,0,sizeof(a))   //memset_assign.
#define absent(a,x)      a.find(x)==a.end()        //absent
#define sz(a)          (int)((a).size())
#define IOS            ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define da(p,t)          p t;cin>>t             //define_assign.
#define da2(p,t1,t2)     p t1,t2; cin>>t1>>t2
#define da3(p,t1,t2,t3)  p t1,t2,t3;cin>>t1>>t2>>t3
#define darr(p,arr,n)   p arr[n]; rl(i,n)cin>>arr[i]
#define edl             "\n"
#define lb(a,x) lower_bound(all(a),x)
#define ub(a,x) upper_bound(all(a),x)
#define make_unique(vec) sort(all(vec));vec.erase(unique(all(vec)),vec.end())
#define lll __uint128_t
const double eps=1e-10;
const float epsf=1e-6;
//const ll  mod=998244353;
//const ll mod = 1e9+7;
using namespace std;
inline ll __gcd(ll a,ll b){if(a==0||b==0){return max(a,b);}ll tempa,tempb;while(1){if(a%b==0)return b;else{tempa=a;tempb=b;a=tempb;b=tempa%tempb;}}}
inline int compfloat(float& x, float& y){if(fabs(x - y) < epsf)return 0;else if(x-y>0)return 1;return -1;}
inline int compdouble(double x, double y){if(fabs(x - y) < eps)return 0;else if(x-y>0)return 1;else return -1;}
bool prime(ll k){for(ll i=2;i*i<=k;i++)if(k%i==0){return false;}return true;}
void pdash(int n=1){for(int i=0;i<n;i++){for(int j=0;j<30;j++){cout<<"-";}cout<<edl;}}
ll power(ll x,ll y){ll result =1;while(y>0){if(y&1){result=(result*x);}y=y>>1;x=(x*x);}return result;}
ll power(ll x,ll y,ll z){ll result =1;x=x%z;while(y>0){if(y&1){result=(result*x)%z;}y=y>>1;x=(x*x)%z;}return result;}
ll modInverse(ll n,ll p){return power(n, p-2, p);}
ll nCrF(ll n,ll r,ll p){if(r==0)return 1;ll f[n+1];f[0]=1;for(ll i=1;i<=n;i++)f[i]=f[i-1]*i%p;
    return (f[n]*modInverse(f[r],p)%p*modInverse(f[n-r],p)%p)%p;}
void cordinate_compression(vector<int>& v){vector<int >p=v;make_unique(p);rl(i,sz(v))v[i]=(int)(lb(p,v[i])-p.begin());}
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
void solve()
{
    int n;cin>>n;
    ll k;cin>>k;
    ll temp = (k-n%k)%k;
    ll maxi=(temp+n-1)/n;
    cout<<maxi+1<<edl;

}
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
int main(){
        IOS;  //comment it when you are in interactive
        int t=1;
        cin>>t;
    while(t--){
        solve();}
}
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/*
       * do smth instead of nothing and stay organized
       * don't listen music while programming.
*/
/*
  stuff you should look for
    * int overflow, array bounds
    * special cases (n=1?), set tle
*/



### Prompt

In cpp, your task is to solve 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 PIE acos(-1)
#define sa(arr,n)    cout<<"-------arr---------\n";rl(i,n)cout<<i<<" "<<arr[i]<<"\n";cout<<"\n";
#define sac(arr,j,n) cout<<"-------arr--------\n";rlc(i,j,n)cout<<i<<" "<<arr[i]<<"\n";cout<<"\n";
#define rlc(i,a,n)     for(int i=a;i<n;i++)   //run_loop
#define rl(i,n)        for(int i=0;i<n;i++)  //run_loop2
#define farr(arr,n)   rl(i,n)cin>>arr[i]   //fill_array
#define farrc(arr,a,n) rlc(i,a,n)cin>>arr[i]
#define ll             long long int
#define ld             long double
#define pb             push_back
#define all(v)         v.begin(),v.end()
#define eb             emplace_back
#define memas(a)        memset(a,-1,sizeof(a))   //memset_assign.
#define mema0(a)        memset(a,0,sizeof(a))   //memset_assign.
#define absent(a,x)      a.find(x)==a.end()        //absent
#define sz(a)          (int)((a).size())
#define IOS            ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define da(p,t)          p t;cin>>t             //define_assign.
#define da2(p,t1,t2)     p t1,t2; cin>>t1>>t2
#define da3(p,t1,t2,t3)  p t1,t2,t3;cin>>t1>>t2>>t3
#define darr(p,arr,n)   p arr[n]; rl(i,n)cin>>arr[i]
#define edl             "\n"
#define lb(a,x) lower_bound(all(a),x)
#define ub(a,x) upper_bound(all(a),x)
#define make_unique(vec) sort(all(vec));vec.erase(unique(all(vec)),vec.end())
#define lll __uint128_t
const double eps=1e-10;
const float epsf=1e-6;
//const ll  mod=998244353;
//const ll mod = 1e9+7;
using namespace std;
inline ll __gcd(ll a,ll b){if(a==0||b==0){return max(a,b);}ll tempa,tempb;while(1){if(a%b==0)return b;else{tempa=a;tempb=b;a=tempb;b=tempa%tempb;}}}
inline int compfloat(float& x, float& y){if(fabs(x - y) < epsf)return 0;else if(x-y>0)return 1;return -1;}
inline int compdouble(double x, double y){if(fabs(x - y) < eps)return 0;else if(x-y>0)return 1;else return -1;}
bool prime(ll k){for(ll i=2;i*i<=k;i++)if(k%i==0){return false;}return true;}
void pdash(int n=1){for(int i=0;i<n;i++){for(int j=0;j<30;j++){cout<<"-";}cout<<edl;}}
ll power(ll x,ll y){ll result =1;while(y>0){if(y&1){result=(result*x);}y=y>>1;x=(x*x);}return result;}
ll power(ll x,ll y,ll z){ll result =1;x=x%z;while(y>0){if(y&1){result=(result*x)%z;}y=y>>1;x=(x*x)%z;}return result;}
ll modInverse(ll n,ll p){return power(n, p-2, p);}
ll nCrF(ll n,ll r,ll p){if(r==0)return 1;ll f[n+1];f[0]=1;for(ll i=1;i<=n;i++)f[i]=f[i-1]*i%p;
    return (f[n]*modInverse(f[r],p)%p*modInverse(f[n-r],p)%p)%p;}
void cordinate_compression(vector<int>& v){vector<int >p=v;make_unique(p);rl(i,sz(v))v[i]=(int)(lb(p,v[i])-p.begin());}
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
void solve()
{
    int n;cin>>n;
    ll k;cin>>k;
    ll temp = (k-n%k)%k;
    ll maxi=(temp+n-1)/n;
    cout<<maxi+1<<edl;

}
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
int main(){
        IOS;  //comment it when you are in interactive
        int t=1;
        cin>>t;
    while(t--){
        solve();}
}
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/*
       * do smth instead of nothing and stay organized
       * don't listen music while programming.
*/
/*
  stuff you should look for
    * int overflow, array bounds
    * special cases (n=1?), set tle
*/



```