#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <cstring>
#include <numeric>
#include <algorithm>
#include <functional>
#include <array>
#include <map>
#include <queue>
#include <limits.h>
#include <set>
#include <stack>
#include <random>
#include <complex>
#include <unordered_map>
#define rep(i,s,n) for(int i = (s); (n) > i; i++)
#define REP(i,n) rep(i,0,n)
#define RANGE(x,a,b) ((a) <= (x) && (x) < (b))
#define DUPLE(a,b,c,d) (RANGE(a,c,d) || RANGE(b,c,d) || RANGE(c,a,b) || RANGE(d,a,b))
#define INCLU(a,b,c,d) (RANGE(a,c,d) && (b,c,d))
#define PW(x) ((x)*(x))
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define MODU 1000000007LL
#define bitcheck(a,b)   ((a >> b) & 1)
#define bitset(a,b)      ( a |= (1 << b))
#define bitunset(a,b)    (a &= ~(1 << b))
#define MP(a,b) make_pair((a),(b))
#define Manh(a,b) (abs((a).first-(b).first) + abs((a).second - ((b).second))
#define pritnf printf
#define scnaf scanf
#define itn int
#define PI 3.141592653589


#define izryt bool
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
template<typename A, size_t N, typename T>
void Fill(A(&array)[N], const T &val) {
	std::fill((T*)array, (T*)(array + N), val);
}
pll Dir[8] = { //移動
	{ 0 ,1 },{ -1 ,0 },{ 1 ,0 },{ 0 ,-1 },
	{ 1 ,1 },{ 1 ,-1 },{ -1 ,1 },{ -1 ,-1 }
};

//[a, b)
#define Getsum(ar, a,b) (ar[b] - ar[a])
#define INF 10000000000000000LL

struct Edge {
	ll from, to;
	ll w;
};
signed main(void) {

	ll a, b, x;
	cin >> a >> b >> x;
	ll k = (x - a) / (a - b);
	if (x - a < 0) k = -1;
	k++;
	k %= MODU;

	cout << ((k *( a%MODU))%MODU + x - (k * ((a - b)%MODU))%MODU)%MODU << endl;

	return 0;
}

### 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 <cstdio>
#include <vector>
#include <cmath>
#include <cstring>
#include <numeric>
#include <algorithm>
#include <functional>
#include <array>
#include <map>
#include <queue>
#include <limits.h>
#include <set>
#include <stack>
#include <random>
#include <complex>
#include <unordered_map>
#define rep(i,s,n) for(int i = (s); (n) > i; i++)
#define REP(i,n) rep(i,0,n)
#define RANGE(x,a,b) ((a) <= (x) && (x) < (b))
#define DUPLE(a,b,c,d) (RANGE(a,c,d) || RANGE(b,c,d) || RANGE(c,a,b) || RANGE(d,a,b))
#define INCLU(a,b,c,d) (RANGE(a,c,d) && (b,c,d))
#define PW(x) ((x)*(x))
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define MODU 1000000007LL
#define bitcheck(a,b)   ((a >> b) & 1)
#define bitset(a,b)      ( a |= (1 << b))
#define bitunset(a,b)    (a &= ~(1 << b))
#define MP(a,b) make_pair((a),(b))
#define Manh(a,b) (abs((a).first-(b).first) + abs((a).second - ((b).second))
#define pritnf printf
#define scnaf scanf
#define itn int
#define PI 3.141592653589


#define izryt bool
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
template<typename A, size_t N, typename T>
void Fill(A(&array)[N], const T &val) {
	std::fill((T*)array, (T*)(array + N), val);
}
pll Dir[8] = { //移動
	{ 0 ,1 },{ -1 ,0 },{ 1 ,0 },{ 0 ,-1 },
	{ 1 ,1 },{ 1 ,-1 },{ -1 ,1 },{ -1 ,-1 }
};

//[a, b)
#define Getsum(ar, a,b) (ar[b] - ar[a])
#define INF 10000000000000000LL

struct Edge {
	ll from, to;
	ll w;
};
signed main(void) {

	ll a, b, x;
	cin >> a >> b >> x;
	ll k = (x - a) / (a - b);
	if (x - a < 0) k = -1;
	k++;
	k %= MODU;

	cout << ((k *( a%MODU))%MODU + x - (k * ((a - b)%MODU))%MODU)%MODU << endl;

	return 0;
}

```