#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int MOD2 = 998244353;
vector<int> bases;
const long long int LLINF = 1ll << 60;
const char en = '\n';
void yes() {
  cout << "YES" << en;
  exit(0);
}
void no() {
  cout << "NO" << en;
  exit(0);
}
template <class T>
void prVec(vector<T> w) {
  cout << w.size() << endl;
  for (int i = 0; i < int(w.size()) - 1; ++i) cout << w[i] << ' ';
  cout << w[w.size() - 1] << endl;
}
long long int raand() {
  long long int a = rand();
  a *= RAND_MAX;
  a += rand();
  return a % (MOD - 7);
}
string to_upper(string a) {
  for (int i = 0; i < (int)a.size(); ++i)
    if (a[i] >= 'a' && a[i] <= 'z') a[i] -= 'a' - 'A';
  return a;
}
string to_lower(string a) {
  for (int i = 0; i < (int)a.size(); ++i)
    if (a[i] >= 'A' && a[i] <= 'Z') a[i] += 'a' - 'A';
  return a;
}
long long int sti(string a) {
  long long int k = 0;
  for (int i = 0; i < (int)a.size(); ++i) {
    k *= 10;
    k += a[i] - '0';
  }
  return k;
}
string its(long long int k) {
  if (k == 0) return "0";
  string a;
  while (k) {
    a.push_back((k % 10) + '0');
    k /= 10;
  }
  reverse(a.begin(), a.end());
  return a;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T, class K>
pair<T, K> mp(T a, K b) {
  return make_pair(a, b);
}
long long int mult(long long int a, long long int b) { return (a * b) % MOD; }
long long int pot(long long int n, long long int k) {
  if (k == 0) return 1;
  long long int a = pot(n, k / 2);
  a = mult(a, a);
  if (k % 2)
    return mult(a, n);
  else
    return a;
}
long long int divide(long long int a, long long int b) {
  return mult(a, pot(b, MOD - 2));
}
long long int sub(long long int a, long long int b) {
  return (a - b + MOD) % MOD;
}
long long int add(long long int a, long long int b) { return (a + b) % MOD; }
bool prime(long long int a) {
  if (a == 1) return 0;
  for (int i = 2; i <= round(sqrt(a)); ++i) {
    if (a % i == 0) return 0;
  }
  return 1;
}
pair<int, int> has(string first) {
  long long int h1 = 0, h2 = 0;
  first = to_lower(first);
  for (char a : first) {
    h1 *= bases[0];
    h1 += a - 'a';
    h1 %= MOD;
    h2 *= bases[1];
    h2 += a - 'a';
    h2 %= MOD2;
  }
  return make_pair(h1, h2);
}
const int N = 300010;
int t;
long long int a, b, c;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int main() {
  ios_base::sync_with_stdio(false);
  srand((long long)new char);
  for (int i = 0; i < 10; ++i) bases.push_back(raand() + 13893829);
  cin >> t;
  while (t--) {
    cin >> a >> b >> c;
    if (a > b) swap(a, b);
    if (c < b)
      cout << -1 << en;
    else {
      long long int rez = 0;
      if (a % 2 != b % 2) {
        if (b % 2 == c % 2)
          --b;
        else
          --a;
        --c;
      }
      rez += c;
      if (b % 2 != c % 2) rez -= 2;
      cout << rez << en;
    }
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

Mikhail walks on a Cartesian plane. He starts at the point (0, 0), and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point (0, 0), he can go to any of the following points in one move: 

  * (1, 0); 
  * (1, 1); 
  * (0, 1); 
  * (-1, 1); 
  * (-1, 0); 
  * (-1, -1); 
  * (0, -1); 
  * (1, -1). 



If Mikhail goes from the point (x1, y1) to the point (x2, y2) in one move, and x1 ≠ x2 and y1 ≠ y2, then such a move is called a diagonal move.

Mikhail has q queries. For the i-th query Mikhail's target is to go to the point (n_i, m_i) from the point (0, 0) in exactly k_i moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point (0, 0) to the point (n_i, m_i) in k_i moves.

Note that Mikhail can visit any point any number of times (even the destination point!).

Input

The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of queries.

Then q lines follow. The i-th of these q lines contains three integers n_i, m_i and k_i (1 ≤ n_i, m_i, k_i ≤ 10^{18}) — x-coordinate of the destination point of the query, y-coordinate of the destination point of the query and the number of moves in the query, correspondingly.

Output

Print q integers. The i-th integer should be equal to -1 if Mikhail cannot go from the point (0, 0) to the point (n_i, m_i) in exactly k_i moves described above. Otherwise the i-th integer should be equal to the the maximum number of diagonal moves among all possible movements.

Example

Input

3
2 2 3
4 3 7
10 1 9


Output

1
6
-1

Note

One of the possible answers to the first test case: (0, 0) → (1, 0) → (1, 1) → (2, 2).

One of the possible answers to the second test case: (0, 0) → (0, 1) → (1, 2) → (0, 3) → (1, 4) → (2, 3) → (3, 2) → (4, 3).

In the third test case Mikhail cannot reach the point (10, 1) in 9 moves.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int MOD2 = 998244353;
vector<int> bases;
const long long int LLINF = 1ll << 60;
const char en = '\n';
void yes() {
  cout << "YES" << en;
  exit(0);
}
void no() {
  cout << "NO" << en;
  exit(0);
}
template <class T>
void prVec(vector<T> w) {
  cout << w.size() << endl;
  for (int i = 0; i < int(w.size()) - 1; ++i) cout << w[i] << ' ';
  cout << w[w.size() - 1] << endl;
}
long long int raand() {
  long long int a = rand();
  a *= RAND_MAX;
  a += rand();
  return a % (MOD - 7);
}
string to_upper(string a) {
  for (int i = 0; i < (int)a.size(); ++i)
    if (a[i] >= 'a' && a[i] <= 'z') a[i] -= 'a' - 'A';
  return a;
}
string to_lower(string a) {
  for (int i = 0; i < (int)a.size(); ++i)
    if (a[i] >= 'A' && a[i] <= 'Z') a[i] += 'a' - 'A';
  return a;
}
long long int sti(string a) {
  long long int k = 0;
  for (int i = 0; i < (int)a.size(); ++i) {
    k *= 10;
    k += a[i] - '0';
  }
  return k;
}
string its(long long int k) {
  if (k == 0) return "0";
  string a;
  while (k) {
    a.push_back((k % 10) + '0');
    k /= 10;
  }
  reverse(a.begin(), a.end());
  return a;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T, class K>
pair<T, K> mp(T a, K b) {
  return make_pair(a, b);
}
long long int mult(long long int a, long long int b) { return (a * b) % MOD; }
long long int pot(long long int n, long long int k) {
  if (k == 0) return 1;
  long long int a = pot(n, k / 2);
  a = mult(a, a);
  if (k % 2)
    return mult(a, n);
  else
    return a;
}
long long int divide(long long int a, long long int b) {
  return mult(a, pot(b, MOD - 2));
}
long long int sub(long long int a, long long int b) {
  return (a - b + MOD) % MOD;
}
long long int add(long long int a, long long int b) { return (a + b) % MOD; }
bool prime(long long int a) {
  if (a == 1) return 0;
  for (int i = 2; i <= round(sqrt(a)); ++i) {
    if (a % i == 0) return 0;
  }
  return 1;
}
pair<int, int> has(string first) {
  long long int h1 = 0, h2 = 0;
  first = to_lower(first);
  for (char a : first) {
    h1 *= bases[0];
    h1 += a - 'a';
    h1 %= MOD;
    h2 *= bases[1];
    h2 += a - 'a';
    h2 %= MOD2;
  }
  return make_pair(h1, h2);
}
const int N = 300010;
int t;
long long int a, b, c;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int main() {
  ios_base::sync_with_stdio(false);
  srand((long long)new char);
  for (int i = 0; i < 10; ++i) bases.push_back(raand() + 13893829);
  cin >> t;
  while (t--) {
    cin >> a >> b >> c;
    if (a > b) swap(a, b);
    if (c < b)
      cout << -1 << en;
    else {
      long long int rez = 0;
      if (a % 2 != b % 2) {
        if (b % 2 == c % 2)
          --b;
        else
          --a;
        --c;
      }
      rez += c;
      if (b % 2 != c % 2) rez -= 2;
      cout << rez << en;
    }
  }
}

```