#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using lld = long double;
template <typename t>
inline void print(const t &v) {
  for (const auto &i : v) {
    std::cout << i << " ";
  }
  std::cout << '\n';
}
template <typename t>
inline void printt(const t &v) {
  for (const auto &i : v) {
    std::cout << i.first << " " << i.second << '\n';
  }
  std::cout << '\n';
}
ll bin_pow(ll a, ll b) {
  if (b == 0) return 1;
  long long res = bin_pow(a, b / 2);
  if (b % 2)
    return res * res * a;
  else
    return res * res;
}
int digit(ll i) { return i > 0 ? (int)log10((double)i) + 1 : 1; }
vector<bool> isPrime(2000000, true);
void make_seive() {
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i = 2; i <= sqrt(2000000); i++) {
    if (isPrime[i]) {
      for (int j = i * 2; j <= 2000000; j += i) {
        isPrime[j] = false;
      }
    }
  }
}
void solve() {
  int a, b;
  cin >> a >> b;
  if (a < b) {
    int diff = b - a;
    if (diff == 0)
      cout << 0;
    else if (diff & 1)
      cout << 1;
    else
      cout << 2;
  } else {
    int diff = a - b;
    if (diff == 0)
      cout << 0;
    else if (diff & 1)
      cout << 2;
    else
      cout << 1;
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please formulate a cpp solution 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;
using ll = long long int;
using lld = long double;
template <typename t>
inline void print(const t &v) {
  for (const auto &i : v) {
    std::cout << i << " ";
  }
  std::cout << '\n';
}
template <typename t>
inline void printt(const t &v) {
  for (const auto &i : v) {
    std::cout << i.first << " " << i.second << '\n';
  }
  std::cout << '\n';
}
ll bin_pow(ll a, ll b) {
  if (b == 0) return 1;
  long long res = bin_pow(a, b / 2);
  if (b % 2)
    return res * res * a;
  else
    return res * res;
}
int digit(ll i) { return i > 0 ? (int)log10((double)i) + 1 : 1; }
vector<bool> isPrime(2000000, true);
void make_seive() {
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i = 2; i <= sqrt(2000000); i++) {
    if (isPrime[i]) {
      for (int j = i * 2; j <= 2000000; j += i) {
        isPrime[j] = false;
      }
    }
  }
}
void solve() {
  int a, b;
  cin >> a >> b;
  if (a < b) {
    int diff = b - a;
    if (diff == 0)
      cout << 0;
    else if (diff & 1)
      cout << 1;
    else
      cout << 2;
  } else {
    int diff = a - b;
    if (diff == 0)
      cout << 0;
    else if (diff & 1)
      cout << 2;
    else
      cout << 1;
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) solve();
}

```