Yet Another Two Integers Problem
You are given two integers $$$a$$$ and $$$b$$$.In one move, you can choose some integer $$$k$$$ from $$$1$$$ to $$$10$$$ and add it to $$$a$$$ or subtract it from $$$a$$$. In other words, you choose an integer $$$k \in [1; 10]$$$ and perform $$$a := a + k$$$ or $$$a := a - k$$$. You may use different values of $$$k$$$ in different moves.Your task is to find the minimum number of moves required to obtain $$$b$$$ from $$$a$$$.You have to answer $$$t$$$ independent test cases.
In the first test case of the example, you don't need to do anything.In the second test case of the example, the following sequence of moves can be applied: $$$13 \rightarrow 23 \rightarrow 32 \rightarrow 42$$$ (add $$$10$$$, add $$$9$$$, add $$$10$$$).In the third test case of the example, the following sequence of moves can be applied: $$$18 \rightarrow 10 \rightarrow 4$$$ (subtract $$$8$$$, subtract $$$6$$$).
The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. Then $$$t$$$ test cases follow.The only line of the test case contains two integers $$$a$$$ and $$$b$$$ ($$$1 \le a, b \le 10^9$$$).
For each test case, print the answer: the minimum number of moves required to obtain $$$b$$$ from $$$a$$$.
def main():
    for s in [*open(0)][1:]:
        a = [*map(int, s.split())]
        b = abs(a[0] - a[1])
        if b % 10:
            print(b // 10 + 1)
        else:
            print(b // 10)
main()
t = int(input())
min_ls = []
for i in range(t):
    (a, b) = (int(j) for j in input().split())
    min_ls.append(str(abs(a - b) // 10 + (1 if abs(a - b) % 10 != 0 else 0)))
print('\n'.join(min_ls))
import math
t = int(input())
res = []
while t:
    t -= 1
    inp = str(input()).split(' ')
    (a, b) = (int(inp[0]), int(inp[1]))
    res.append(math.ceil(abs(a - b) / 10))
for r in res:
    print(r)