#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool sortbysec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
string decToBin(long long int n) {
  string s = "0";
  if (n == 0) return s;
  s = "";
  while (n > 0) {
    s += (n % 2) + '0';
    n = n / 2;
  }
  reverse(s.begin(), s.end());
  return s;
}
long long int binToDec(string s) {
  long long int ans = 0;
  reverse(s.begin(), s.end());
  for (__typeof(s.size()) i = (0) - ((0) > (s.size()));
       i != (s.size()) - ((0) > (s.size())); i += 1 - 2 * ((0) > (s.size()))) {
    ans += pow(2, i) * (s[i] - '0');
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int q;
  cin >> q;
  while (q--) {
    long long int a, b;
    cin >> a >> b;
    if (a == b) cout << 0 << endl;
    if (a < b) {
      if (a % 2 == b % 2)
        cout << 2 << endl;
      else
        cout << 1 << endl;
    }
    if (a > b) {
      if (a % 2 == b % 2)
        cout << 1 << endl;
      else
        cout << 2 << endl;
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool sortbysec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
string decToBin(long long int n) {
  string s = "0";
  if (n == 0) return s;
  s = "";
  while (n > 0) {
    s += (n % 2) + '0';
    n = n / 2;
  }
  reverse(s.begin(), s.end());
  return s;
}
long long int binToDec(string s) {
  long long int ans = 0;
  reverse(s.begin(), s.end());
  for (__typeof(s.size()) i = (0) - ((0) > (s.size()));
       i != (s.size()) - ((0) > (s.size())); i += 1 - 2 * ((0) > (s.size()))) {
    ans += pow(2, i) * (s[i] - '0');
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int q;
  cin >> q;
  while (q--) {
    long long int a, b;
    cin >> a >> b;
    if (a == b) cout << 0 << endl;
    if (a < b) {
      if (a % 2 == b % 2)
        cout << 2 << endl;
      else
        cout << 1 << endl;
    }
    if (a > b) {
      if (a % 2 == b % 2)
        cout << 1 << endl;
      else
        cout << 2 << endl;
    }
  }
  return 0;
}

```