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 main() {
  int t;
  cin >> t;
  while (t) {
    int a, b;
    cin >> a >> b;
    if (a == 0 || b == 0) {
      cout << 0 << endl;
    } else {
      int maxof = max(a, b);
      int minof = min(a, b);
      if (maxof >= 2 * minof) {
        cout << minof << endl;
      } else {
        int sum = 0;
        int c = 0;
        int f1 = maxof / 3;
        int f2 = minof / 3;
        int k = min(f1, f2);
        sum += 2 * k;
        maxof = maxof - k * 3;
        minof = minof - k * 3;
        while (maxof > 0 && minof > 0) {
          if (c == 0 && maxof >= 2 && minof >= 1) {
            sum += 1;
            maxof -= 2;
            minof -= 1;
            c = 1;
          } else {
            c = 1;
          }
          if (c == 1 && maxof >= 1 && minof >= 2) {
            sum += 1;
            maxof -= 1;
            minof -= 2;
            c = 0;
          } else {
            c = 0;
          }
          if (maxof == 1 && minof == 1) {
            break;
          }
        }
        cout << sum << endl;
      }
    }
    t--;
  }
  return 0;
}
