#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep1(i, n) for (int i = 1; i < (n); i++)
#define all(v) (v).begin(), (v).end()
using namespace std;
typedef long long LL;
const int INF = 999999999;
const LL mod = 1000000007;

int main(){
    LL a,b,x;
    cin >> a >> b >> x;
    if(x < a){
        cout << x%mod << endl;
        return 0;
    }
    LL t = ((x-b) / (a-b)) % mod;
    cout << (x%mod + t*(b%mod))%mod << endl;
}


### Prompt

Your challenge is to write a cpp solution to the following problem:

Problem

If you bring an empty bottle of $ a $ milk, you can exchange it for a new bottle of $ b $ milk.
How many bottles of milk can Mr. Kawabayashi, who initially has a bottle of $ x $ milk, drink? Output the remainder after dividing by $ 1000000007 $.

Constraints

The input satisfies the following conditions.

* $ 1 \ leq a \ leq 10 ^ {15} $
* $ 0 \ leq b \ lt a $
* $ 0 \ leq x \ leq 10 ^ {15} $

Input

The input is given in the following format.


$ a $ $ b $ $ x $


Three integers $ a $, $ b $, $ x $ are given, separated by spaces.

Output

Print the answer on one line.

Examples

Input

3 1 5


Output

7


Input

3 2 5


Output

11


Input

82 69 64


Output

64


Input

316250877917604 316250877917599 681260158257385


Output

62687552

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep1(i, n) for (int i = 1; i < (n); i++)
#define all(v) (v).begin(), (v).end()
using namespace std;
typedef long long LL;
const int INF = 999999999;
const LL mod = 1000000007;

int main(){
    LL a,b,x;
    cin >> a >> b >> x;
    if(x < a){
        cout << x%mod << endl;
        return 0;
    }
    LL t = ((x-b) / (a-b)) % mod;
    cout << (x%mod + t*(b%mod))%mod << endl;
}


```