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;
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
bool compare(long long int a, long long int b) { return a >= b; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int test;
  cin >> test;
  while (test--) {
    long long int a, b;
    cin >> a >> b;
    long long int ans = 0, count = 0, cnt = 0;
    bool flag = 1;
    long long int x = min(a, b), y = max(a, b);
    if (x * 2 <= y)
      ans = x;
    else {
      ans = y - x;
      x = x - ans;
      y = y - ans * 2;
      count = x / 3;
      count *= 2;
      cnt = x % 3;
      long long int dif = y - x + cnt;
      if (cnt == 1 && dif > 1) {
        count++;
      } else if (cnt == 2) {
        if (dif > 3)
          count += 2;
        else if (dif > 1)
          count++;
      }
      ans += count;
    }
    cout << ans << endl;
  }
}
