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;
long long power(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res = (res * x);
    }
    n >>= 1;
    if (n != 0) {
      x = (x * x);
    }
  }
  return res;
}
long long powerMOD(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res = (res * x) % 1000000007;
    }
    n >>= 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
bool mycompare(pair<long long, long long> &a, pair<long long, long long> &b) {
  if (a.first == b.first) {
    return a.second < b.second;
  }
  return a.first > b.first;
}
bool mycompare1(pair<int, int> &a, pair<int, int> &b) {
  return a.second < b.second;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  long long i, m, j, k, n, t, x, l, c, ans;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> n >> k;
    if (n > k) {
      swap(n, k);
    }
    if (2 * n <= k) {
      cout << n << "\n";
      continue;
    }
    cout << (n + k) / 3 << "\n";
  }
}
