#include <math.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <array>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

using namespace std;

#define rep(i, n) for (int i = 0; i < (n); i++)
#define loop(i, a, n) for (int i = a; i < (n); i++)
#define all(in) in.begin(), in.end()
const int INF = 1e9;
const long long MOD = 1e9 + 7;

int main() {
    long long a, b, x;
    cin >> a >> b >> x;
    long long ans = x + (max(x - b, 0ll) / (a - b)) % MOD * (b % MOD);
    cout << ans % MOD << endl;
    return 0;
}

### Prompt

Generate 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 <math.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <array>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

using namespace std;

#define rep(i, n) for (int i = 0; i < (n); i++)
#define loop(i, a, n) for (int i = a; i < (n); i++)
#define all(in) in.begin(), in.end()
const int INF = 1e9;
const long long MOD = 1e9 + 7;

int main() {
    long long a, b, x;
    cin >> a >> b >> x;
    long long ans = x + (max(x - b, 0ll) / (a - b)) % MOD * (b % MOD);
    cout << ans % MOD << endl;
    return 0;
}

```