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;
const int INF = 2147483647;
const double pi = 3.1415926535897932384626;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
int t, a, b;
int main() {
  t = read();
  while (t--) {
    a = read(), b = read();
    int ans;
    if (a == 0 || b == 0)
      ans = 0;
    else {
      if (a * 2 <= b)
        ans = b / 2;
      else if (a >= b * 2)
        ans = a / 2;
      else {
        ans = 2 * min(a / 3, b / 3);
        a %= 3, b %= 3;
        if (a >= 2 && b)
          ++ans;
        else if (a && b >= 2)
          ++ans;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
