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 Mod = 1e9 + 7;
bool isPowerOfTwo(int x) { return x && (!(x & (x - 1))); }
long long int pow(long long int x, long long int y, long long int mod) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = (ans * x) % mod;
    y /= 2;
    x = (x * x) % mod;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t, a, b, temp, ans;
  cin >> t;
  while (t--) {
    cin >> a >> b;
    if (a > b) swap(a, b);
    if (a == 0 || b == 1) {
      cout << 0 << endl;
      continue;
    }
    ans = 0;
    if (b >= 2 * a) {
      cout << a << endl;
      continue;
    }
    while (a >= 1 && b >= 2) {
      temp = b - a;
      if (temp == 1 || temp == 0) break;
      temp /= 2;
      if (temp > a) break;
      b -= temp * 2;
      a -= temp;
      if (a == 0 || b == 1) break;
      ans += temp;
    }
    if (a == b) {
      temp = a / 3;
      ans += 2 * temp;
      temp = a % 3;
      if (temp == 2) {
        ans += 1;
      }
    } else if (b == a + 1) {
      if (a >= 1 && b >= 2) {
        ans++;
        b -= 2;
        a--;
        temp = a / 3;
        ans += 2 * temp;
        temp = a % 3;
        if (temp == 2) {
          ans += 1;
        }
      }
    }
    cout << ans << endl;
  }
}
