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;
int prime[100000];
void seive(int n) {
  int i, j;
  for (i = 2; i <= sqrt(n); i++) {
    if (prime[i] == 0) {
      for (j = i * i; j <= n; j += i) {
        prime[j] = 1;
      }
    }
  }
}
int main() {
  long long int i, j, k, x = 0, l = 101, c = 0, b = 0, t, r, a = 0, m, n;
  string s, s1;
  map<string, int> mp;
  vector<int> vect, vect2, vect1;
  cin >> t;
  while (t--) {
    l = 0;
    cin >> a >> b;
    m = a / 2;
    n = b / 2;
    x = a % 2;
    int y = b % 2;
    if (a == 0 || b == 0)
      cout << "0" << endl;
    else if (max(m, n) >= min(a, b))
      cout << min(a, b) << endl;
    else {
      cout << (a + b) / 3 << endl;
    }
  }
}
