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>
#pragma warning(disable : 4996)
using namespace std;
int t, ans;
void bf(int a, int b, int val) {
  if ((a <= 1 && b <= 1) || b == 0) {
    ans = max(ans, val);
    return;
  }
  int v1 = min(a / 2, b);
  bf(max(a - v1 * 2, b - v1), min(a - v1 * 2, b - v1), val + v1);
  if (a < 3 || b < 3) return;
  int v2 = min(a / 3, b / 3);
  bf(max(a - v2 * 3, b - v2 * 3), min(a - v2 * 3, b - v2 * 3), val + v2 * 2);
  return;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    int a, b;
    scanf("%d %d", &a, &b);
    ans = 0;
    if (a < b) swap(a, b);
    bf(a, b, 0);
    printf("%d\n", ans);
  }
  return 0;
}
