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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long binexp(long long x, long long n, long long m) {
  if (!n)
    return 1;
  else if (n & 1)
    return (x % m) * binexp((x % m) * (x % m), n / 2, m) % m;
  else if (!(n & 1))
    return binexp((x % m) * (x % m), n / 2, m) % m;
  return 0;
}
int32_t solve(long long _, long long _t) {
  long long a, b;
  cin >> a >> b;
  if (a == 0 || b == 0) return cout << 0 << '\n', 0;
  if (a < b) swap(a, b);
  long long x = a - b;
  long long ans = x;
  if (x > b) ans = min({a / 2, b});
  a -= 2ll * x;
  b -= x;
  if (a > 0 && b > 0) ans += 2 * max(a, 0ll) / 3;
  cout << ans << '\n';
  return 0;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  long long _ = 1;
  cin >> _;
  long long _t = _;
  while (_--) solve(_, _t);
  return 0;
}
