Problem:
Score : 100 points

Problem Statement

You went shopping to buy cakes and donuts with X yen (the currency of Japan).

First, you bought one cake for A yen at a cake shop.
Then, you bought as many donuts as possible for B yen each, at a donut shop.

How much do you have left after shopping?

Constraints

1 \leq A, B \leq 1 000

A + B \leq X \leq 10 000

X, A and B are integers.

Input

Input is given from Standard Input in the following format:

X
A
B

Output

Print the amount you have left after shopping.

Sample Input 1

1234
150
100

Sample Output 1

84

You have 1234 - 150 = 1084 yen left after buying a cake.
With this amount, you can buy 10 donuts, after which you have 84 yen left.

Sample Input 2

1000
108
108

Sample Output 2

28

Sample Input 3

579
123
456

Sample Output 3

0

Sample Input 4

7477
549
593

Sample Output 4

405

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
 
#define rep(i, n) for (int i = (int)(0); i < (int)(n); ++i)
#define reps(i, n) for (int i = (int)(1); i <= (int)(n); ++i)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define rreps(i, n) for (int i = ((int)(n)); i > 0; i--)
#define irep(i, m, n) for (int i = (int)(m); i < (int)(n); ++i)
#define ireps(i, m, n) for (int i = (int)(m); i <= (int)(n); ++i)
#define FOR(e, c) for (auto &e : c)
#define SORT(v, n) sort(v, v + n);
#define vsort(v) sort(v.begin(), v.end());
#define rvisort(v) sort(v.begin(), v.end(), greater<int>());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cout(d) cout<<d<<endl;
#define coutd(d) cout<<std::setprecision(10)<<d<<endl;
#define cinline(n) getline(cin,n);
 
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using vpii = vector<pii>;
using vs = vector<string>;
using vd = vector<double>;
using ul = unsigned long;
 
template<class T, class C> void chmax(T& a, C b){ a>b?:a=b; }
template<class T, class C> void chmin(T& a, C b){ a<b?:a=b; }
 
const int mod=1e9+7;
 
struct mint {
  ll x;
  mint(ll x=0):x(x%mod){}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
};
 
mint c[4005][4005];
void init() {
  c[0][0] = 1;
  for (int i = 0; i <= 4000; i++) {
    for (int j = 0; j <= i; j++) {
      c[i+1][j] += c[i][j];
      c[i+1][j+1] += c[i][j];
    }
  }
}
mint comb(int n, int k) {
  return c[n][k];
}
 
bool IsPrime(int num)
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false;
 
    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            return false;
        }
    }
 
    return true;
}
 
map<ll, ll> primeFact(ll n) {
  map<ll, ll> res;
  for (ll i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      res[i]++;
      n /= i;
    }
  }
  if (n != 1) res[n]++;
  return res;
}
 
void printv(vi& v){
  rep(i,v.size()) cout << v[i] << " ";
  cout << endl;
}
void printv(vll& v){
  rep(i,v.size()) cout << v[i] << " ";
  cout << endl;
}
int vtotal(vi& v){
  int total=0;
  rep(i,v.size()) total+=v[i];
  return total;
}
 
int main()
{
  //init(); // mint初期化
  cin.tie( 0 );
  ios::sync_with_stdio( false );
  int x,a,b;
  cin>>x>>a>>b;
  int ans=(x-a)%b;
  cout(ans);
  return 0;
}