#include <bits/stdc++.h>
  
using namespace std;
  
#define rep(i,n) REP(i,0,n)
#define REP(i,s,e) for(int i=(s); i<(int)(e); i++)
#define repr(i, n) REPR(i, n, 0)
#define REPR(i, s, e) for(int i=(int)(s-1); i>=(int)(e); i--)
#define pb push_back
#define all(r) r.begin(),r.end()
#define rall(r) r.rbegin(),r.rend()
#define fi first
#define se second
  
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
 
const int INF = 1e9;
const ll MOD = 1e9 + 7;
double EPS = 1e-8;

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

### Prompt

Create a solution in Cpp for 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>
  
using namespace std;
  
#define rep(i,n) REP(i,0,n)
#define REP(i,s,e) for(int i=(s); i<(int)(e); i++)
#define repr(i, n) REPR(i, n, 0)
#define REPR(i, s, e) for(int i=(int)(s-1); i>=(int)(e); i--)
#define pb push_back
#define all(r) r.begin(),r.end()
#define rall(r) r.rbegin(),r.rend()
#define fi first
#define se second
  
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
 
const int INF = 1e9;
const ll MOD = 1e9 + 7;
double EPS = 1e-8;

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

```