#include<bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    cin>>t;

    while(t--)
    {
        long long a,b;
        cin>>a>>b;

        if(b<10)
            cout<<b-a<<endl;

        else
        {
            long long ans=b-a,x,y;
            long long c=10;
            while(1)
            {
                x=a/c;
                y=b/c;

                if(x==0 && y==0)
                    break;
                ans+=(y-x);

                c=c*10;

            }

            cout<<ans<<endl;
        }


    }



    return 0;
}


### Prompt

Develop a solution in cpp to the problem described below:

You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.

For example: 

  * if l=909, then adding one will result in 910 and 2 digits will be changed; 
  * if you add one to l=9, the result will be 10 and 2 digits will also be changed; 
  * if you add one to l=489999, the result will be 490000 and 5 digits will be changed. 



Changed digits always form a suffix of the result written in the decimal system.

Output the total number of changed digits, if you want to get r from l, adding 1 each time.

Input

The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).

Output

For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.

Example

Input


4
1 9
9 10
10 20
1 1000000000


Output


8
2
11
1111111110

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    cin>>t;

    while(t--)
    {
        long long a,b;
        cin>>a>>b;

        if(b<10)
            cout<<b-a<<endl;

        else
        {
            long long ans=b-a,x,y;
            long long c=10;
            while(1)
            {
                x=a/c;
                y=b/c;

                if(x==0 && y==0)
                    break;
                ans+=(y-x);

                c=c*10;

            }

            cout<<ans<<endl;
        }


    }



    return 0;
}


```