
#include <bits/stdc++.h>

#define ll long long
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define vi vector<int>
#define pii pair<int, int>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;
template<typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;

void solve() {
    ll l, r;
    cin >> l >> r;
    ll ans = 0;
    rep(i, 0, 10) {
        ans += r - l;
        l /= 10;
        r /= 10;
    }
    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int te;
    cin >> te;
    while(te--) solve();
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>

#define ll long long
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define vi vector<int>
#define pii pair<int, int>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;
template<typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;

void solve() {
    ll l, r;
    cin >> l >> r;
    ll ans = 0;
    rep(i, 0, 10) {
        ans += r - l;
        l /= 10;
        r /= 10;
    }
    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int te;
    cin >> te;
    while(te--) solve();
}
```