For all positive integers (x\ge 1) and (k\ge 2), let (v_k(x!)) denote the number of trailing zeros in the base-(k) representation of (x! = x\cdot (x-1)\cdot \ldots \cdot 1). Formally, (v_k(x!)) is defined as the largest integer (i) such that (k^i) divides (x!). For a prime number (p), we can calculate (v_p(x!) = \sum\limits_{j=1}^\infty \left\lfloor \frac{x}{p^j}\right\rfloor)(^{\text{∗}}). If (k) is not prime, write its prime factorization as (k = \prod p_i^{e_i}), where (p_i) are distinct prime factors and (e_i) are their corresponding exponents. Then, ()v_k(x!) = \min\limits_i \left\lfloor \frac{v_{p_i}(x!)}{e_i}\right\rfloor.() For any two positive integers (a) and (b), and any integer (k\ge 2), the weight of the pair ((a, b)) with respect to (k), denoted by (w_k(a, b)), is defined as ()w_k(a, b) = \begin{cases}\min(v_k(a!), v_k(b!)) & \text{if }v_k(a!)\neq v_k(b!)\text{;}\\10^{100} & \text{otherwise.}\end{cases}() Next, define (f_m(a, b)) as the minimum weight of the pair ((a, b)) with respect to (k), taken over all integers (k) with (2\le k\le m): ()f_m(a, b)=\min\limits_{2\le k\le m}w_k(a, b).() You are given two integers (n) and (m). Your task is to compute the sum of (f_m(x, n)) over all positive integers (x) less than (n): ()\sum_{1\le x\le n - 1} f_m(x, n).() It can be shown that under the given constraints, the result is strictly less than (10^{100}). (^{\text{∗}})(\lfloor y\rfloor) denotes the floor of (y), which is the greatest integer less than or equal to (y). Each test contains multiple test cases. The first line contains the number of test cases (t) ((1 \le t \le 100)). The description of the test cases follows. The first and only line of each test case contains two integers (n) and (m) ((2\le n\le m\le 10^7)) — the parameters of the function