#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <stdlib.h>
#include <string>
#include <utility>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define loop(i, x, n) for (int i = (x); i < (n); i++)
#define all(v) (v).begin(), (v).end()
#define int long long
using namespace std;
const int MOD = 1e9 + 7;
const int INF = 1e18;
template<typename T> void cmax(T &a, T b) { a = max(a, b); }
template<typename T> void cmin(T &a, T b) { a = min(a, b); }

signed main() {
  int a, b, x;
  cin >> a >> b >> x;

  if (a > x) {
    cout << x % MOD << endl;
    return 0;
  }
  int m = (x - a) / (a - b) + 1;
  cout << (int)((x + (b % MOD) * (m % MOD)) % MOD) << endl;
  return 0;
}


### Prompt

Please create a solution in cpp 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 <algorithm>
#include <bitset>
#include <cmath>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <stdlib.h>
#include <string>
#include <utility>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define loop(i, x, n) for (int i = (x); i < (n); i++)
#define all(v) (v).begin(), (v).end()
#define int long long
using namespace std;
const int MOD = 1e9 + 7;
const int INF = 1e18;
template<typename T> void cmax(T &a, T b) { a = max(a, b); }
template<typename T> void cmin(T &a, T b) { a = min(a, b); }

signed main() {
  int a, b, x;
  cin >> a >> b >> x;

  if (a > x) {
    cout << x % MOD << endl;
    return 0;
  }
  int m = (x - a) / (a - b) + 1;
  cout << (int)((x + (b % MOD) * (m % MOD)) % MOD) << endl;
  return 0;
}


```