#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcountll

#define INF 1e16
#define mod 1000000007

ll x,a,b;

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  cin>>a>>b>>x;
  if(x<a){
    cout<<x%mod<<endl;
    return 0;
  }

  ll d=(a-b);
  ll n=(x-a)/d; n++;

  a%=mod; b%=mod; x%=mod; d%=mod; n%=mod;
  ll res=((a*n%mod)+(x-(d*n%mod)+mod))%mod;
  cout<<res<<endl;

  return 0;
}


### Prompt

Construct a CPP code solution to the problem outlined:

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;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcountll

#define INF 1e16
#define mod 1000000007

ll x,a,b;

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  cin>>a>>b>>x;
  if(x<a){
    cout<<x%mod<<endl;
    return 0;
  }

  ll d=(a-b);
  ll n=(x-a)/d; n++;

  a%=mod; b%=mod; x%=mod; d%=mod; n%=mod;
  ll res=((a*n%mod)+(x-(d*n%mod)+mod))%mod;
  cout<<res<<endl;

  return 0;
}


```