//↓template↓

#include "bits/stdc++.h"
using namespace std;
#define Would
#define you
#define all(n)  n.begin(),n.end()
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
const double pi = acos(-1);
const int SIZE = 1 << 17;
int dx[] = { 1,0 }, dy[] = { 0,1 }, alp[30];
long long fac[200005], finv[200005], inv[200005];
vector<long long>dij;
struct edge { long long to, cost; };
vector<vector<edge> >G;

long long mod_pow(long long a, long long b) {
	long long res = 1, mul = a;
	for (int i = 0; i < 31; ++i) {
		if (b >> i & 1) {
			res *= mul;
			res %= MOD;
		}
		mul = (mul * mul) % MOD;
	}

	return res;
}

void addedge(int from, int to, int cost) {
	G[from].push_back({ to,cost });
	G[to].push_back({ from,cost });
}

//↑template↑

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

### Prompt

Develop a solution in CPP to the problem described below:

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
//↓template↓

#include "bits/stdc++.h"
using namespace std;
#define Would
#define you
#define all(n)  n.begin(),n.end()
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
const double pi = acos(-1);
const int SIZE = 1 << 17;
int dx[] = { 1,0 }, dy[] = { 0,1 }, alp[30];
long long fac[200005], finv[200005], inv[200005];
vector<long long>dij;
struct edge { long long to, cost; };
vector<vector<edge> >G;

long long mod_pow(long long a, long long b) {
	long long res = 1, mul = a;
	for (int i = 0; i < 31; ++i) {
		if (b >> i & 1) {
			res *= mul;
			res %= MOD;
		}
		mul = (mul * mul) % MOD;
	}

	return res;
}

void addedge(int from, int to, int cost) {
	G[from].push_back({ to,cost });
	G[to].push_back({ from,cost });
}

//↑template↑

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

```