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 long long int Mod = 1e9 + 7, MX = 1e7, MN = LLONG_MIN;
void Print(long long int a, long long int b = MN, long long int c = MN,
           long long int d = MN) {
  printf("%lld", a);
  if (b != MN) printf(" %lld", b);
  if (c != MN) printf(" %lld", c);
  if (d != MN) printf(" %lld", d);
  puts("");
}
int main() {
  long long int a, b;
  scanf("%*d");
  while (~scanf("%lld %lld", &a, &b)) {
    long long int ans = 0;
    while (a && b && abs(a - b) > 1) {
      if (a > b) swap(a, b);
      long long int d = b - a;
      b -= d;
      if (d & 1) b++, d--;
      if (a - d / 2 < 1) {
        ans += a;
        a = b = 0;
      } else {
        long long int x = d / 2;
        a -= x, ans += x;
      }
    }
    if (a > b) swap(a, b);
    if (a >= 3) {
      long long int x = a - (a % 3);
      a %= 3;
      ans += x / 3 * 2;
      b -= x;
    }
    if (a > b) swap(a, b);
    if (b >= 2 && a >= 1) ans++;
    printf("%lld\n", ans);
  }
  return 0;
}
