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;
const int mod = 1e9 + 7;
const int maxn = 5e4 + 10;
const int maxm = 5e5 + 10;
const long long INF = 1e18;
long long powmod(long long x, long long y) {
  long long t;
  for (t = 1; y; y >>= 1, x = x * x % mod)
    if (y & 1) t = t * x % mod;
  return t;
}
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int n, a, b, ret = 0;
void solve(int x, int y) {
  int ta = a - x, tb = b - y;
  if (ta < 0 || tb < 0) return;
  if ((ta + tb) % 3) return;
  int tp = (ta + tb) / 3;
  if (ta < tp || tb < tp) return;
  ret = max(ret, ta + tb - 2 * tp);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  while (n--) {
    ret = 0;
    cin >> a >> b;
    solve(0, 0);
    solve(0, 1);
    solve(1, 0);
    solve(1, 1);
    cout << ret << "\n";
  }
  return 0;
}
