B. Difference of GCDs

time limit per test 1 second

memory limit per tes t256 megabytes

input standard input

output standard output

You are given three integers n, l, and r. You need to construct an array a1,a2,…,an (l≤ai≤r) such that gcd(i,ai)gcd(i,ai) are all distinct or report there's no solution.

Here gcd(x,y)gcd(x,y) denotes the greatest common divisor (GCD) of integers x and y.

Input

The input consists of multiple test cases. The first line contains a single integer t (1≤t≤10^4) — the number of test cases. The description of the test cases follows.

The first line contains three integers nn, ll, rr (1≤n≤10^5, 1≤l≤r≤10^9).

It is guaranteed that the sum of nn over all test cases does not exceed 105105.

Output

For each test case, if there is no solution, print "NO" (without quotes). You can print letters in any case (upper or lower).

Otherwise, print "YES" (without quotes). In the next line, print n integers a1,a2,…,an — the array you construct.

If there are multiple solutions, you may output any.

Example

input

4
5 1 5
9 1000 2000
10 30 35
1 1000000000 1000000000

output

YES
1 2 3 4 5
YES
1145 1926 1440 1220 1230 1350 1001 1000 1233
NO
YES
1000000000

Note

In the first test case, gcd(1,a1),gcd(2,a2),…,gcd(5,a5) are equal to 1, 2, 3, 4, 5, respectively.

OK, What we know is that we need to find a number (k For a random variable )k*i(i>=1&&i<=n) stay [l,r] Between ,

Then there is k>=l/i If l%i==0, Then we will a[i]==k=l/i.

If l%i!=0, Then we have to think about it , or l>i or l<i, All in all l/i Is a decimal , But we use int type , All it will be forcibly rounded , in other words （int）l/i<(float)l/i, We put k=int(l/i)+1 That's it ,a[i]==k=l/i+1;

if(a[i]>r) It means in [l,r] There is no number that satisfies the meaning of the question , We output NO, Otherwise output YES.

The above sentence means that we are [l,r] Find the smallest i Multiple k）

( for instance l=4,i=4, that k=1,k*i=4;

l=4,i=2, that k=2,k*i=4;

l=5,i=2, that k=3,k*i=6;

l=5,i=4, that k=2,k*i=8;)

The code is as follows ：

#include <iostream>
using namespace std;
int n,l,r;
int main()
{
    int t;
    cin>>t;
    while (t>0)
    {
        cin>>n>>l>>r;
        bool falg=true;
        int data1[100001];
        for(int i=1;i<=n;i++)
        {
            if(l%i==0)
            {
                data1[i]=l;
            }
            else
            {
                if(((l/i)+1)*i>r)                   /* If the minimum multiples exceed r Words , There is no solution */
                {
                    falg=false;
                    break;
                }
                else
                {
                    data1[i]=((l/i)+1)*i;
                }
            }
        }
        if(falg==true)
        {
            cout<<"YES"<<endl;
            for(int i=1;i<=n;i++)
            {
                cout<<data1[i]<<" ";
            }
            cout<<endl;
        }
        else
        {
            cout<<"NO"<<endl;
        }
        t--;
    }
    return 0;
}