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>
bool isFile;
using namespace std;
const int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
void _();
void faster() {
  _();
  if (isFile) {
    freopen(
        "taskA"
        ".in",
        "r", stdin);
    freopen(
        "taskA"
        ".out",
        "w", stdout);
  }
  if (1) {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
}
const long long INF = 9e18;
const int inf = 2e9;
const int MOD = 1e9 + 7;
const long double PI = acos(-1);
const long double EPS = 3E-16;
const int ALPHA = 27;
const int N = (int)2e5 + 5;
const int M = 1234;
void solve() {
  long long n, m;
  cin >> n >> m;
  long long ans = 0, res = 0, q = 0;
  ans = min(n / 2, m) + (n % 2 && (m - min(n / 2, m)) >= 2);
  res = min(n, m / 2) + (m % 2 && (n - min(m / 2, n)) >= 2);
  {
    q = min(n / 3, m / 3);
    n -= q * 3;
    m -= q * 3;
    q *= 2;
    q += max(min(n / 2, m) + (n % 2 && (m - min(n / 2, m)) >= 2),
             min(n, m / 2) + (m % 2 && (n - min(m / 2, n)) >= 2));
  }
  cout << max({ans, res, q}) << endl;
}
void _() {}
int main() {
  faster();
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
