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>
#pragma GCC optimize(3, "Ofast", "inline")
const int maxn = 10010;
const int MaxN = 0x3f3f3f3f;
const int MinN = 0xc0c0c00c;
const int mod = 100000000;
using namespace std;
int x, y;
int check(int n) { return n + min((x - 2 * n), (y - n) / 2); }
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> x >> y;
    int l = 0, r = min(x / 2, y);
    int ans = 0;
    while (l <= r) {
      int mid1 = l + (r - l) / 3;
      int mid2 = r - (r - l) / 3;
      if (l == r) {
        ans = check(l);
        break;
      }
      if (check(mid1) > check(mid2)) {
        ans = check(mid1);
        r = mid2 - 1;
      } else
        l = mid1 + 1;
    }
    cout << ans << endl;
  }
  return 0;
}
