1352C. K-th Not Divisible by n

binary search, math, 1200, https://codeforces.com/problemset/problem/1352/C

You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.

For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 . The 7-th number among them is 10.

Input

The first line contains an integer t ($1\le t\le 1000$) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n ($2\le n\le {10}^9$) and k ($1\le k\le {10}^9$).

Output

For each test case print the k-th positive integer that is not divisible by n.

Example

input

6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1

output

10
15
1999999999
113
1000000001
1

思路：

观察规律

在每一段连续的 n 个整数中：

• 有 n - 1 个数不是 n 的倍数；
• 有 1 个数是 n 的倍数。

---

数学推导法

可以推导出一个直接公式：

cnt = k // (n - 1)
rem = k % (n - 1)

解释：

• 每 n 个数中有 n - 1 个有效。
• 所以经过 cnt 组（每组 n 个数），共计 cnt * (n - 1) 个有效数。

剩余还需要 rem 个。

那么答案是：

x = cnt * n + rem

但要注意一个细节：

• 当 rem == 0 时，说明恰好在某个 n 的倍数前停下； 那就要往前退一个倍数：

  x = cnt * n - 1

t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    cnt = k // (n - 1)
    rem = k % (n - 1)
    if rem == 0:
        print(cnt * n - 1)
    else:
        print(cnt * n + rem)