#include <bits/stdc++.h>
using namespace std;
template <typename Type>
vector<Type> readArray(vector<Type> a, int n) {
  for (int i = 0; i < n; i++) cin >> a[i];
  return a;
}
template <typename Type>
vector<vector<Type>> read2Array(vector<vector<Type>> a, int n, int m) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j];
  return a;
}
template <typename Type>
void printArray(vector<Type> a, int n) {
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
template <typename Type>
void print2Array(vector<vector<Type>> a, int n, int m) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cout << a[i][j] << " ";
    cout << "\n";
  }
}
template <typename Type>
using pq = priority_queue<Type>;
template <typename Type>
using pq_r = priority_queue<Type, vector<Type>, greater<Type>>;
bool prime(int x) {
  if (x == 1) return false;
  if (x == 2) return true;
  for (int i = 2; i <= sqrt(x) + 1; i++)
    if (x % i == 0) return false;
  return true;
}
void solve() {
  int a, b;
  cin >> a >> b;
  if (a == b)
    cout << 0 << "\n";
  else {
    if (b > a) {
      if ((a - b) % 2 != 0)
        cout << 1 << "\n";
      else
        cout << 2 << "\n";
    } else {
      if ((a - b) % 2 == 0)
        cout << 1 << "\n";
      else
        cout << 2 << "\n";
    }
  }
}
int main() {
  int tt = 1;
  cin >> tt;
  while (tt--) solve();
  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;
template <typename Type>
vector<Type> readArray(vector<Type> a, int n) {
  for (int i = 0; i < n; i++) cin >> a[i];
  return a;
}
template <typename Type>
vector<vector<Type>> read2Array(vector<vector<Type>> a, int n, int m) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j];
  return a;
}
template <typename Type>
void printArray(vector<Type> a, int n) {
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
template <typename Type>
void print2Array(vector<vector<Type>> a, int n, int m) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cout << a[i][j] << " ";
    cout << "\n";
  }
}
template <typename Type>
using pq = priority_queue<Type>;
template <typename Type>
using pq_r = priority_queue<Type, vector<Type>, greater<Type>>;
bool prime(int x) {
  if (x == 1) return false;
  if (x == 2) return true;
  for (int i = 2; i <= sqrt(x) + 1; i++)
    if (x % i == 0) return false;
  return true;
}
void solve() {
  int a, b;
  cin >> a >> b;
  if (a == b)
    cout << 0 << "\n";
  else {
    if (b > a) {
      if ((a - b) % 2 != 0)
        cout << 1 << "\n";
      else
        cout << 2 << "\n";
    } else {
      if ((a - b) % 2 == 0)
        cout << 1 << "\n";
      else
        cout << 2 << "\n";
    }
  }
}
int main() {
  int tt = 1;
  cin >> tt;
  while (tt--) solve();
  return 0;
}

```