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 bexpo(int x, int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return bexpo(x * x, n / 2);
  else
    return x * bexpo(x * x, (n - 1) / 2);
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int n, k;
bool check(int mid, int a[], int ind) {
  int cnt = 0, i;
  for (i = 0; i < n; i++) {
    if (mid >= a[i] && cnt % 2 == ind)
      cnt++;
    else if (cnt % 2 != ind)
      cnt++;
  }
  return cnt >= k;
}
int main() {
  int t, i;
  cin >> t;
  while (t--) {
    long long int a, b;
    cin >> a >> b;
    long long int ans = min(a, b);
    ans = min(ans, (a + b) / 3);
    cout << ans << endl;
  }
}
