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;
int calculate_emeralds(long long int *x, long long int *y) {
  long long int result = 0;
  result = min(*x / 2, *y);
  *x -= (2 * result);
  *y -= result;
  return result;
}
int main() {
  int t;
  cin >> t;
  while (t) {
    long long int a, b, emeralds1 = 0, emeralds2 = 0, emeralds3 = 0, copy_a,
                        copy_b;
    cin >> a >> b;
    copy_a = a, copy_b = b;
    emeralds1 += calculate_emeralds(&copy_a, &copy_b);
    emeralds1 += calculate_emeralds(&copy_b, &copy_a);
    copy_a = a, copy_b = b;
    emeralds2 += calculate_emeralds(&copy_b, &copy_a);
    emeralds2 += calculate_emeralds(&copy_a, &copy_b);
    copy_a = a, copy_b = b;
    emeralds3 += (min(copy_a / 3, copy_b / 3) * 2);
    copy_a -= (min(a / 3, b / 3) * 3);
    copy_b -= (min(a / 3, b / 3) * 3);
    a = copy_a, b = copy_b;
    long long int x = calculate_emeralds(&copy_a, &copy_b);
    copy_a = a, copy_b = b;
    long long int y = calculate_emeralds(&copy_b, &copy_a);
    emeralds3 += max(x, y);
    cout << max(max(emeralds1, emeralds2), emeralds3) << endl;
    t--;
  }
}
