#include <bits/stdc++.h>
#define _GLIBCXX_DEBUG
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define Sort(a) sort(a.begin(), a.end())
#define Reve(a) reverse(a.begin(), a.end())
#define PI 3.14159265359
typedef long long ll;
const ll MOD = 1e9+7;

ll gcd(ll a, ll b){
  if(b==0) return a;
  return gcd(b, a%b);
}
/*ll myPow(ll x, ll n, ll m){
  if(n == 0)
    return 1;
  if(n % 2 == 0)
    return myPow(x * x % m, n / 2, m);
  else
    return x * myPow(x, n - 1, m) % m;
}*/

ll lcm(ll a, ll b){
  ll g = gcd(a, b);
  b /= g;
  a *= b;
  return a;
}

//素数判定 O(√n)
bool is_prime(int64_t n){
  for(int64_t i = 2; i*i<=n; i++){
    if(n%i==0) return false;
  }
  return n!=1;
}

//約数の列挙 O(√n)
vector<int64_t> dividor(int64_t n){
  vector<int64_t> res;
  for(int64_t i=1; i*i<=n; i++){
    if(n%i==0){
      res.push_back(i);
      if(i != n/i) res.push_back(i);
    }
  }
  return res;
}

//素因数分解 O(√n)
map<int64_t, int64_t> prime_factor(int64_t n){
  map<int64_t, int64_t> res;
  for(int64_t i=2; i*i<=n; i++){
    while(n%i==0){
      res[i]++;
      n /= i;
    }
  }
  if(n != 1) res[n] = 1;
  return res;
}
struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};

/*int main(){
  int n, m, q; cin >> n >> m >> q;
  vector<int> C(q), D(q), A(m, 1);
  //vector<pair<int, int>> AB(q);
  map<int, tuple<int, int, int>> S;

  vector<int> E(n, -1);*/
  /*rep(i, q) cin >> AB[i].first >> AB[i].second >> C[i] >> D[i];*/
  /*rep(i, q){
    int a, b, c, d;
    cin >> a >> b >> c >> D[i];
    S[d] = make_tuple(a, b, c);
  }
  Sort(D);
  int ans=0;*/
  
  /*rep(i, q){
    int da=D[i];
    int n1 = get<0>(S[da]);
    int n2 = get<1>(S[da]);
    int n3 = get<1>(S[da]);
    if(E[n1-1]==-1&&E[n2-1]==-1) {
      if(n1+n3<=m){
        E[n1-1]=n1;
        E[n2-1]=n1+n3;
      }
      if(n1+n3<)
    }
    else if(E[AB[i].first-1]==-1){

    }*/
  //}    
//}

int main(){
  //cout << 2 << endl;
  ll a, b, n; cin >> a >> b >> n;
  if(n<=1000000000){
    ll ans = 0; ll tmp = 0;
    rep(i, n+1){
      ll g = (a*i)/b;
      ll h = i/b;
      if(tmp<g-a*h) {tmp=g-a*h; ans = i;}
    }
    cout << tmp << endl;
  }
  else {
    //ll n1 = (ll) sqrt(n);
    //cout << n1 << endl;
    /*ll ans = 0; ll tmp = 0;
    for(ll i=n; i>n-b; i--){
      ll g = a*i/b;
      ll h = i/b;
      if(tmp<g-a*h) {tmp=g-a*h; ans = i;}
    }*/
    ll tmp1 = 0;
    if(n>b-1){ll g1 = a*(b-1)/b;
    ll h1 = (b-1)/b;
    tmp1 = g1-a*h1;}
    ll g2 = a*n/b;
    ll h2 = n/b;
    ll tmp2 = g2-a*h2;
    cout << max(tmp1, tmp2) << endl;
    }
  }
  
  




  





Measured memory_bytes: 3624