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;
template <class T>
T sqr(T val) {
  return val * val;
}
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
void solve() {
  long long int a, b;
  cin >> a >> b;
  if (a == 0 || b == 0) {
    cout << 0 << endl;
    return;
  }
  if (a > b) {
    if (a >= b * 2)
      cout << b << endl;
    else if (a > b) {
      int del = b / 2;
      b = b - del;
      cout << del + b / 2 << endl;
    } else
      cout << a / 2 << endl;
  } else {
    if (b >= a * 2)
      cout << a << endl;
    else if (b > a) {
      int del = a / 2;
      a = a - del;
      cout << del + a / 2 << endl;
    } else
      cout << b / 2 << endl;
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  long long TC = 1;
  cin >> TC;
  while (TC--) {
    solve();
  }
}
