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 = (int)1e9 + 5;
const long long LINF = (long long)4e18 + 5;
const int M = 1e9 + 7;
template <typename T>
T gcd(T a, T b) {
  return (b == 0) ? abs(a) : gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    int a, b;
    cin >> a >> b;
    int x = min(a, b);
    int y = max(a, b);
    int ans = 0, a1 = 0;
    int s;
    int x1, y1;
    if (y >= 2 * x) {
      ans = x;
    } else {
      x1 = x;
      y1 = y;
      a1 = y1 / 2;
      x1 -= a1;
      y1 -= a1 * 2;
      if (x1 > 1 && y1 == 1) {
        a1++;
      }
      ans += (x / 3) * 2;
      s = (x / 3) * 3;
      x -= s;
      y -= s;
      if (y >= 2 * x) {
        ans += x;
      } else {
        ans += (y / 2);
      }
      ans = max(ans, a1);
    }
    cout << ans << '\n';
  }
}
