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;
unsigned long long min(unsigned long long x, unsigned long long y) {
  if (x < y) return x;
  return y;
}
unsigned long long max(unsigned long long x, unsigned long long y) {
  if (x < y) return y;
  return x;
}
long long min(long long x, long long y) {
  if (x < y) return x;
  return y;
}
long long max(long long x, long long y) {
  if (x < y) return y;
  return x;
}
double min(double x, double y) {
  if (x < y) return x;
  return y;
}
double max(double x, double y) {
  if (x < y) return y;
  return x;
}
unsigned long long gcd(unsigned long long x, unsigned long long y) {
  if (!x) return y;
  if (!y) return x;
  if (x > y) swap(x, y);
  return gcd(x, y % x);
}
unsigned long long inv(unsigned long long a, unsigned long long c) {
  if (a == 1) {
    return 1;
  }
  return ((c - (c / a)) * inv(c % a, c)) % c;
}
long long a, b;
int check(long long k) {
  if ((k > a) || (k > b)) return 0;
  long long ux = a - k, lx = max((long long)2 * k - b, 0), uy = b - k,
            ly = max((long long)2 * k - a, 0);
  return (lx <= ux) && (ly <= uy);
}
long long bs() {
  long long l = 0, r = 1e9 + 7;
  while (l < r - 1) {
    long long k = l + (r - l) / 2;
    if (check(k)) {
      l = k;
    } else {
      r = k;
    }
  }
  if (check(r)) return r;
  return l;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    cin >> a >> b;
    cout << bs() << "\n";
  }
  return 0;
}
