#include <bits/stdc++.h>
using namespace std;
void mego() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void modBigNumber(string num, long long m) {
  vector<int> vec;
  long long mod = 0;
  bool zeroflag = 0;
  for (int i = 0; i < num.size(); i++) {
    int digit = num[i] - '0';
    mod = mod * 10 + digit;
    int quo = mod / m;
    if (!zeroflag && quo > 0) {
      cout << quo;
      zeroflag = 1;
    } else if (zeroflag)
      cout << quo;
    mod = mod % m;
  }
  if (!zeroflag) {
    cout << 0;
    return;
  }
  return;
}
int fch(char ch, string s) {
  int idx = 0;
  for (; idx < s.size(); idx++)
    if (s[idx] == ch) break;
  if (idx == s.size())
    return -1;
  else
    return idx;
}
int main() {
  mego();
  long long a, b, c = 0, d, n, t, m, k;
  string s, s1;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    b -= a;
    if (b == 0)
      cout << 0 << "\n";
    else if (b < 0 && !(-b % 2))
      cout << 1 << "\n";
    else if (b > 0 && (b & 1))
      cout << 1 << "\n";
    else
      cout << 2 << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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;
void mego() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void modBigNumber(string num, long long m) {
  vector<int> vec;
  long long mod = 0;
  bool zeroflag = 0;
  for (int i = 0; i < num.size(); i++) {
    int digit = num[i] - '0';
    mod = mod * 10 + digit;
    int quo = mod / m;
    if (!zeroflag && quo > 0) {
      cout << quo;
      zeroflag = 1;
    } else if (zeroflag)
      cout << quo;
    mod = mod % m;
  }
  if (!zeroflag) {
    cout << 0;
    return;
  }
  return;
}
int fch(char ch, string s) {
  int idx = 0;
  for (; idx < s.size(); idx++)
    if (s[idx] == ch) break;
  if (idx == s.size())
    return -1;
  else
    return idx;
}
int main() {
  mego();
  long long a, b, c = 0, d, n, t, m, k;
  string s, s1;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    b -= a;
    if (b == 0)
      cout << 0 << "\n";
    else if (b < 0 && !(-b % 2))
      cout << 1 << "\n";
    else if (b > 0 && (b & 1))
      cout << 1 << "\n";
    else
      cout << 2 << "\n";
  }
  return 0;
}

```