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;
long long pow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
void solve(long long t) {
  while (t--) {
    long long i, j, k, l, m, n, a, b, z, ans;
    cin >> a >> b;
    if (a >= b) {
      z = (a / 2);
      ans = min(b, z);
    } else {
      z = (b / 2);
      ans = min(a, z);
    }
    if (a < 1 || b < 1 || (a == 1 && b <= 1) || (a <= 1 && b == 1)) {
      cout << 0 << "\n";
    } else if (a >= 3 && b >= 3) {
      if (a >= b) {
        if (a >= (2 * b))
          cout << max(ans, b) << "\n";
        else {
          long long temp = b % 3;
          if (!temp) {
            cout << max(ans, 2 * (b / 3)) << "\n";
          } else {
            if (temp == 1) {
              a = a % 3;
              if (a >= 2)
                cout << max(ans, (2 * ((b / 3))) + 1) << "\n";
              else
                cout << max(ans, 2 * (b / 3)) << "\n";
            } else {
              cout << max(ans, (2 * ((b / 3))) + 1) << "\n";
            }
          }
        }
      } else {
        if (b >= (2 * a))
          cout << max(ans, a) << "\n";
        else {
          long long temp = a % 3;
          if (!temp) {
            cout << max(ans, 2 * (a / 3)) << "\n";
          } else {
            if (temp == 1) {
              b = b % 3;
              if (b >= 2)
                cout << max(ans, (2 * ((a / 3))) + 1) << "\n";
              else
                cout << max(ans, 2 * (a / 3)) << "\n";
            } else
              cout << max(ans, (2 * ((a / 3))) + 1) << "\n";
          }
        }
      }
    } else {
      cout << 1 << "\n";
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  solve(t);
  return 0;
}
