Soon after the Chunga-Changa island was discovered, it started to acquire some forms of civilization and even market economy. A new currency arose, colloquially called "chizhik". One has to pay in chizhiks to buy a coconut now.

Sasha and Masha are about to buy some coconuts which are sold at price z chizhiks per coconut. Sasha has x chizhiks, Masha has y chizhiks. Each girl will buy as many coconuts as she can using only her money. This way each girl will buy an integer non-negative number of coconuts.

The girls discussed their plans and found that the total number of coconuts they buy can increase (or decrease) if one of them gives several chizhiks to the other girl. The chizhiks can't be split in parts, so the girls can only exchange with integer number of chizhiks.

Consider the following example. Suppose Sasha has 5 chizhiks, Masha has 4 chizhiks, and the price for one coconut be 3 chizhiks. If the girls don't exchange with chizhiks, they will buy 1 + 1 = 2 coconuts. However, if, for example, Masha gives Sasha one chizhik, then Sasha will have 6 chizhiks, Masha will have 3 chizhiks, and the girls will buy 2 + 1 = 3 coconuts. 

It is not that easy to live on the island now, so Sasha and Mash want to exchange with chizhiks in such a way that they will buy the maximum possible number of coconuts. Nobody wants to have a debt, so among all possible ways to buy the maximum possible number of coconuts find such a way that minimizes the number of chizhiks one girl gives to the other (it is not important who will be the person giving the chizhiks).

Input

The first line contains three integers x, y and z (0 ≤ x, y ≤ 10^{18}, 1 ≤ z ≤ 10^{18}) — the number of chizhics Sasha has, the number of chizhics Masha has and the price of a coconut. 

Output

Print two integers: the maximum possible number of coconuts the girls can buy and the minimum number of chizhiks one girl has to give to the other.

Examples

Input


5 4 3


Output


3 1


Input


6 8 2


Output


7 0

Note

The first example is described in the statement. In the second example the optimal solution is to dot exchange any chizhiks. The girls will buy 3 + 4 = 7 coconuts.
#include <bits/stdc++.h>
using namespace std;
const int dx44[6] = {0, -1, -1, 1, 1};
const int dy44[6] = {0, -1, 1, -1, 1};
char dir_box[] = {'j', 'N', 'S', 'W', 'E'};
char dir_man[] = {'j', 'n', 's', 'w', 'e'};
const int dx4[6] = {0, 1, 0, -1, 0};
const int dy4[6] = {0, 0, 1, 0, -1};
const int dx8[9] = {0, -1, 0, 1, 0, 1, 1, -1, -1};
const int dy8[9] = {0, 0, 1, 0, -1, 1, -1, 1, -1};
const int dx82[9] = {0, -1, -1, 1, 1, 2, 2, -2, -2};
const int dy82[9] = {0, 2, -2, 2, -2, 1, -1, 1, -1};
long long po(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  } else {
    return gcd(b % a, a);
  }
}
void extgcd(long long a, long long b, long long &d, long long &x,
            long long &y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    extgcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long inverse(long long a, long long n) {
  long long d, x, y;
  extgcd(a, n, d, x, y);
  return d == 1 ? (x + n) % n : -1;
}
const int maxn = 1e6 + 5;
const double PI = acos(-1.0);
const long long mod = 1e9 + 7;
const long long mod1 = 1e8 + 7;
const long long mod2 = 1e9 + 7;
void solve() {
  long long x, y, z;
  cin >> x >> y >> z;
  long long ans = (x + y) / z;
  long long ans1 = x / z;
  long long ans2 = y / z;
  long long pri1 = x / z * z;
  long long pri2 = y / z * z;
  long long minn = 1e18;
  if (x % z >= (z - y % z)) {
    minn = z - y % z;
  }
  if (y % z >= (z - x % z)) {
    minn = min(minn, z - x % z);
  }
  if (minn == 1e18) {
    cout << (x + y) / z << " " << 0 << endl;
  } else {
    cout << (x + y) / z << " " << minn << endl;
  }
}
int main() {
  solve();
  return 0;
}
