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 INF = (1LL << 61) - 1;
const long long MOD = 1000000007;
const long long MAX_N = 500010;
long long a, b, c, d, e, f, h, x, y, z, p, q, n, t, r, k, w, l, ans, i, j, u, v;
string S, T;
vector<long long> A, B;
void solve() {
  cin >> a >> b;
  if (a < b) swap(a, b);
  k = min(a - b, b);
  a -= k * 2;
  b -= k;
  if (b == 0) {
    cout << k << "\n";
    return;
  }
  if (a % 3 == 2) {
    cout << k + (a / 3) * 2 + 1 << "\n";
  } else {
    cout << k + (a / 3) * 2 << "\n";
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    solve();
  }
}
