#include<iostream>
#include<iomanip>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<math.h>
#include<functional>
#include<bitset>
using namespace std;
using ll = long long;
using ld = long double;
using pint = pair<int, int>;
using pll = pair<ll, ll>;
#define MOD 1000000007LL
#define INF 1000000000LL
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(int)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v)  sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll a, b, x;
	cin >> a >> b >> x;
	if(x < a) {
		cout << x % MOD << endl;
		return 0;
	}
	ll ans = a % MOD;
	x -= a;
	ll c = (a - b);
	ans += (((x / c) % MOD) * (a % MOD)) % MOD;
	ans += b + ((x % c) % MOD);
	cout << ans % MOD << endl;
}
/* --------------------------------------- */


### Prompt

Your task is to create 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<iostream>
#include<iomanip>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<math.h>
#include<functional>
#include<bitset>
using namespace std;
using ll = long long;
using ld = long double;
using pint = pair<int, int>;
using pll = pair<ll, ll>;
#define MOD 1000000007LL
#define INF 1000000000LL
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(int)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v)  sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll a, b, x;
	cin >> a >> b >> x;
	if(x < a) {
		cout << x % MOD << endl;
		return 0;
	}
	ll ans = a % MOD;
	x -= a;
	ll c = (a - b);
	ans += (((x / c) % MOD) * (a % MOD)) % MOD;
	ans += b + ((x % c) % MOD);
	cout << ans % MOD << endl;
}
/* --------------------------------------- */


```