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 = 998244353;
int pw(int n, int e) {
  int res = 1;
  while (e) {
    if (e % 2) res = ((long long)res * n) % mod;
    n = ((long long)n * n) % mod;
    e >>= 1;
  }
  return res;
}
int inv(int n) { return pw(n, mod - 2); }
int bn(int n, int k) {
  if (k > n) return 0;
  if (k == 0 || n == k) return 1;
  return ((long long)(n % mod) *
          (((long long)bn(n - 1, k - 1) * inv(k)) % mod)) %
         mod;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int a, b;
    cin >> a >> b;
    if (b > a) swap(a, b);
    if (a >= 2 * b)
      cout << b << "\n";
    else
      cout << a - b + ((2 * b - a) / 3) * 2 + ((2 * b - a) % 3 == 2 ? 1 : 0)
           << "\n";
  }
  return 0;
}
