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 = 1e9 + 5;
long long power(long long a, long long b, long long c) {
  long long res = 1;
  while (b) {
    if (b & 1LL) res = (res * a) % c;
    a = (a * a) % c;
    b /= 2;
  }
  return res;
}
void solve() {
  vector<long long> a(2);
  for (long long &x : a) cin >> x;
  sort(a.begin(), a.end());
  if (a[1] >= 2 * a[0]) {
    cout << a[0] << endl;
    return;
  }
  long long ans = min(a[0] / 3, a[1] / 3);
  a[0] -= ans * 3;
  a[1] -= ans * 3;
  ans = ans * 2;
  if (a[1] >= 2 * a[0])
    ans += a[0];
  else if (a[0] >= 1 && a[1] >= 2)
    ans++;
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
