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>
int main() {
  int a, b, n, i;
  int U[1000];
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &a, &b);
    if ((a + b) / 3 > a || (a + b) / 3 > b) {
      if (a < b) {
        U[i] = a;
      } else {
        U[i] = b;
      }
    } else {
      U[i] = (a + b) / 3;
    }
  }
  printf("\n");
  for (i = 0; i < n; i++) {
    printf("%d\n", U[i]);
  }
  return 0;
}
