Polycarp plays a well-known computer game (we won't mention its name). In this game, he can craft tools of two types — shovels and swords. To craft a shovel, Polycarp spends two sticks and one diamond; to craft a sword, Polycarp spends two diamonds and one stick.

Each tool can be sold for exactly one emerald. How many emeralds can Polycarp earn, if he has a sticks and b diamonds?

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

The only line of each test case contains two integers a and b (0 ≤ a, b ≤ 10^9) — the number of sticks and the number of diamonds, respectively.

Output

For each test case print one integer — the maximum number of emeralds Polycarp can earn.

Example

Input


4
4 4
1000000000 0
7 15
8 7


Output


2
0
7
5

Note

In the first test case Polycarp can earn two emeralds as follows: craft one sword and one shovel.

In the second test case Polycarp does not have any diamonds, so he cannot craft anything.
#include <bits/stdc++.h>
using namespace std;
long long mini(long long a, long long b) {
  if (a > b) {
    return b;
  } else {
    return a;
  }
}
long long maxi(long long a, long long b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t, n, a, b, c, d, m;
  cin >> t;
  while (t--) {
    cin >> a >> b;
    if (a == b && a % 3 != 0 && a >= 2) {
      cout << a / 2 << endl;
    } else if (a == b && a % 3 == 0 && a >= 3) {
      cout << a / 2 + 1 << endl;
    } else if ((a == 1 && b == 2) || (b == 1 && a == 2)) {
      cout << 1 << endl;
    } else if ((a < 1) || (b < 1) || (a < 1 && b <= 2) || (b < 1 && a <= 2) ||
               (a <= 1 && b < 2) || (b <= 1 && a < 2)) {
      cout << 0 << endl;
    } else if (a == b && a % 2 == 1 && a > 1) {
      cout << a / 2 + 1;
    } else {
      c = 0;
      d = abs(a - b);
      if (d >= mini(a, b)) {
        cout << mini(a, b) << endl;
      } else {
        m = mini(a, b);
        n = maxi(a, b);
        n = n - 2 * d;
        m = m - d;
        if (m == n && m >= 2 && m % 3 == 0) {
          cout << d + m / 2 + 1 << endl;
        } else if (m == n && m >= 2 && m % 3 != 0) {
          cout << d + m / 2 << endl;
        } else {
          cout << d << endl;
        }
      }
    }
  }
  return 0;
}
