Valera had two bags of potatoes, the first of these bags contains x (x ≥ 1) potatoes, and the second — y (y ≥ 1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains x potatoes) Valera lost. Valera remembers that the total amount of potatoes (x + y) in the two bags, firstly, was not gerater than n, and, secondly, was divisible by k.

Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.

Input

The first line of input contains three integers y, k, n (1 ≤ y, k, n ≤ 109; <image> ≤ 105).

Output

Print the list of whitespace-separated integers — all possible values of x in ascending order. You should print each possible value of x exactly once.

If there are no such values of x print a single integer -1.

Examples

Input

10 1 10


Output

-1


Input

10 6 40


Output

2 8 14 20 26 
#include <bits/stdc++.h>
const long long inf = 1000000000000000000;
using namespace std;
bool prime[1000005];
long long fact[1000005];
long long gcd(long long a, long long b) {
  if (a < b) gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long x = (a / gcd(a, b)) * b;
  return x;
}
long long power(long long x, long long n, long long m = inf) {
  if (x == 0) return 0;
  if (n == 0) return 1;
  long long ans;
  if (n % 2 == 0) {
    long long smallAns = power(x, n / 2, m);
    ans = (smallAns * smallAns) % m;
  } else {
    long long smallAns = power(x, n - 1, m);
    ans = x;
    ans = (ans * smallAns) % m;
  }
  return (ans + m) % m;
}
long long modInverse(long long n, long long p) { return power(n, p - 2, p); }
long long nCr(long long n, long long r, long long p = LLONG_MAX) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (fact[n] * modInverse(fact[r], p) % p * modInverse(fact[n - r], p) %
          p) %
         p;
}
void fillFact() {
  fact[0] = 1;
  for (int i = 1; i < 1000005; i += 1) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
  }
}
void fillPrime() {
  long long i, j, n = 1000001;
  for (int i = 0; i < n; i += 1) {
    prime[i] = 1;
  }
  prime[0] = 0;
  prime[1] = 0;
  for (i = 2; i * i <= n; i++) {
    if (prime[i]) {
      for (j = i * i; j <= n; j += i) {
        prime[j] = 0;
      }
    }
  }
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long kadane(long long arr[], int n) {
  long long dp[n];
  dp[0] = arr[0];
  long long maxm = dp[0];
  for (int i = 1; i < n; i += 1) {
    if (dp[i - 1] > 0)
      dp[i] = arr[i] + dp[i - 1];
    else
      dp[i] = arr[i];
    maxm = max(maxm, dp[i]);
  }
  return maxm;
}
bool compare(pair<int, int> p1, pair<int, int> p2) {
  return p1.second < p2.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    int y, k, n;
    cin >> y >> k >> n;
    int div = (y / k);
    div += 1;
    div *= k;
    if (div > n) {
      cout << -1 << '\n';
      exit(0);
    }
    while (div <= n) {
      cout << div - y << " ";
      div += k;
    }
    cout << '\n';
  }
  return 0;
}
