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 <typename T>
void input(vector<T> &arr, long long int n) {
  T temp;
  for (long long int i = 0; i < n; i++) cin >> temp, arr.push_back(temp);
}
template <typename T>
void output(vector<T> arr) {
  for (auto x : arr) cout << x << " ";
  cout << endl;
}
template <typename T>
void input_set(set<T> &arr, long long int n) {
  T temp;
  for (long long int i = 0; i < n; i++) cin >> temp, arr.insert(temp);
}
long long int mul(long long int a, long long int b) {
  return (a % 1000000007 * b % 1000000007) % 1000000007;
}
long long int power(long long int num, long long int base) {
  long long int ans = 1;
  while (base > 0) {
    if (base & 1) ans = mul(ans, num);
    num = mul(num, num);
    base >>= 1;
  }
  return ans;
}
void solve() {
  long long int second, d;
  cin >> second >> d;
  long long int x = max(second, d);
  long long int y = min(second, d);
  long long int m = 2 * y - x;
  if (m < 0) {
    cout << max(min(x / 2, y), min(y / 2, second)) << endl;
  } else {
    if (m % 3 == 0) {
      m = m / 3;
      cout << y - 2 * m + m << endl;
    } else {
      cout << max(max(m / 3 + 1, min(x / 2, y)), min(y / 2, second)) << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int testcases;
  cin >> testcases;
  while (testcases--) {
    solve();
  }
}
