#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
long long mod = 1e9 + 7;
long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fe(long long base, long long exp) {
  long long ans = 1;
  while (exp) {
    if (exp & 1) ans = mul(ans, base);
    base = mul(base, base);
    exp >>= 1;
  }
  return ans;
}
long long fact(long long n) {
  long long res = 1;
  for (int i = 1; i <= n; i++) {
    res = res * 1ll * i % mod;
  }
  return res;
}
long long modin(long long a) { return fe(a, mod - 2); }
long long nCr(long long a, long long b) {
  if (a < b) return 0;
  return mul(fact(a), mul(modin(fact(b)), modin(fact(a - b))));
}
void solve() {
  int a, b;
  cin >> a >> b;
  if (a == b)
    cout << 0 << "\n";
  else if (a > b) {
    if ((a - b) % 2)
      cout << 2 << "\n";
    else
      cout << 1 << "\n";
  } else {
    if ((b - a) % 2)
      cout << 1 << "\n";
    else
      cout << 2 << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
       << "\n";
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

You are given two positive integers a and b.

In one move, you can change a in the following way:

  * Choose any positive odd integer x (x > 0) and replace a with a+x; 
  * choose any positive even integer y (y > 0) and replace a with a-y. 



You can perform as many such operations as you want. You can choose the same numbers x and y in different moves.

Your task is to find the minimum number of moves required to obtain b from a. It is guaranteed that you can always obtain b from a.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.

Then t test cases follow. Each test case is given as two space-separated integers a and b (1 ≤ a, b ≤ 10^9).

Output

For each test case, print the answer — the minimum number of moves required to obtain b from a if you can perform any number of moves described in the problem statement. It is guaranteed that you can always obtain b from a.

Example

Input


5
2 3
10 10
2 4
7 4
9 3


Output


1
0
2
2
1

Note

In the first test case, you can just add 1.

In the second test case, you don't need to do anything.

In the third test case, you can add 1 two times.

In the fourth test case, you can subtract 4 and add 1.

In the fifth test case, you can just subtract 6.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
long long mod = 1e9 + 7;
long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fe(long long base, long long exp) {
  long long ans = 1;
  while (exp) {
    if (exp & 1) ans = mul(ans, base);
    base = mul(base, base);
    exp >>= 1;
  }
  return ans;
}
long long fact(long long n) {
  long long res = 1;
  for (int i = 1; i <= n; i++) {
    res = res * 1ll * i % mod;
  }
  return res;
}
long long modin(long long a) { return fe(a, mod - 2); }
long long nCr(long long a, long long b) {
  if (a < b) return 0;
  return mul(fact(a), mul(modin(fact(b)), modin(fact(a - b))));
}
void solve() {
  int a, b;
  cin >> a >> b;
  if (a == b)
    cout << 0 << "\n";
  else if (a > b) {
    if ((a - b) % 2)
      cout << 2 << "\n";
    else
      cout << 1 << "\n";
  } else {
    if ((b - a) % 2)
      cout << 1 << "\n";
    else
      cout << 2 << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
       << "\n";
  return 0;
}

```