#include <bits/stdc++.h>
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; c > '9' || c < '0'; c = getchar()) {
    if (c == '-') f = -1;
  }
  for (; c >= '0' && c <= '9'; c = getchar()) {
    x = x * 10 + c - '0';
  }
  return x * f;
}
const int mod = 998244353;
std::vector<long long> divisor;
long long fac[100005] = {1}, inv[100005] = {1};
inline void divide(long long D) {
  for (long long i = 2; i * i <= D; i++) {
    if (D % i == 0) {
      divisor.push_back(i);
      for (; D % i == 0; D /= i)
        ;
    }
  }
  if (D > 1) divisor.push_back(D);
}
inline long long q_pow(long long x, int t) {
  long long s = 1;
  for (; t; t >>= 1) {
    if (t & 1) s = s * x % mod;
    x = x * x % mod;
  }
  return s;
}
inline long long count(long long u, long long v) {
  long long sum = 0, facinv = 1;
  for (auto i : divisor) {
    int cntu = 0, cntv = 0;
    for (; u % i == 0; u /= i) cntu++;
    for (; v % i == 0; v /= i) cntv++;
    if (cntu || cntv) {
      int dif = abs(cntu - cntv);
      facinv = facinv * inv[dif] % mod;
      sum += dif;
    }
  }
  return fac[sum] * facinv % mod;
}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
inline void work() {
  long long D = read();
  divide(D);
  for (int i = 1; i <= 1000; i++) fac[i] = fac[i - 1] * i % mod;
  inv[1000] = q_pow(fac[1000], mod - 2);
  for (int i = 999; i; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  for (int q = read(); q; q--) {
    long long u = read(), v = read(), g = gcd(u, v);
    printf("%lld\n", count(u, g) * count(v, g) % mod);
  }
}
int main() {
  work();
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

You are given a positive integer D. Let's build the following graph from it: 

  * each vertex is a divisor of D (not necessarily prime, 1 and D itself are also included); 
  * two vertices x and y (x > y) have an undirected edge between them if x is divisible by y and \frac x y is a prime; 
  * the weight of an edge is the number of divisors of x that are not divisors of y. 



For example, here is the graph for D=12: 

<image>

Edge (4,12) has weight 3 because 12 has divisors [1,2,3,4,6,12] and 4 has divisors [1,2,4]. Thus, there are 3 divisors of 12 that are not divisors of 4 — [3,6,12].

There is no edge between 3 and 2 because 3 is not divisible by 2. There is no edge between 12 and 3 because 12/3=4 is not a prime.

Let the length of the path between some vertices v and u in the graph be the total weight of edges on it. For example, path [(1, 2), (2, 6), (6, 12), (12, 4), (4, 2), (2, 6)] has length 1+2+2+3+1+2=11. The empty path has length 0.

So the shortest path between two vertices v and u is the path that has the minimal possible length.

Two paths a and b are different if there is either a different number of edges in them or there is a position i such that a_i and b_i are different edges.

You are given q queries of the following form: 

  * v u — calculate the number of the shortest paths between vertices v and u. 



The answer for each query might be large so print it modulo 998244353.

Input

The first line contains a single integer D (1 ≤ D ≤ 10^{15}) — the number the graph is built from.

The second line contains a single integer q (1 ≤ q ≤ 3 ⋅ 10^5) — the number of queries.

Each of the next q lines contains two integers v and u (1 ≤ v, u ≤ D). It is guaranteed that D is divisible by both v and u (both v and u are divisors of D).

Output

Print q integers — for each query output the number of the shortest paths between the two given vertices modulo 998244353.

Examples

Input


12
3
4 4
12 1
3 4


Output


1
3
1


Input


1
1
1 1


Output


1


Input


288807105787200
4
46 482955026400
12556830686400 897
414 12556830686400
4443186242880 325


Output


547558588
277147129
457421435
702277623

Note

In the first example: 

  * The first query is only the empty path — length 0; 
  * The second query are paths [(12, 4), (4, 2), (2, 1)] (length 3+1+1=5), [(12, 6), (6, 2), (2, 1)] (length 2+2+1=5) and [(12, 6), (6, 3), (3, 1)] (length 2+2+1=5). 
  * The third query is only the path [(3, 1), (1, 2), (2, 4)] (length 1+1+1=3). 

### Response

```cpp
#include <bits/stdc++.h>
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; c > '9' || c < '0'; c = getchar()) {
    if (c == '-') f = -1;
  }
  for (; c >= '0' && c <= '9'; c = getchar()) {
    x = x * 10 + c - '0';
  }
  return x * f;
}
const int mod = 998244353;
std::vector<long long> divisor;
long long fac[100005] = {1}, inv[100005] = {1};
inline void divide(long long D) {
  for (long long i = 2; i * i <= D; i++) {
    if (D % i == 0) {
      divisor.push_back(i);
      for (; D % i == 0; D /= i)
        ;
    }
  }
  if (D > 1) divisor.push_back(D);
}
inline long long q_pow(long long x, int t) {
  long long s = 1;
  for (; t; t >>= 1) {
    if (t & 1) s = s * x % mod;
    x = x * x % mod;
  }
  return s;
}
inline long long count(long long u, long long v) {
  long long sum = 0, facinv = 1;
  for (auto i : divisor) {
    int cntu = 0, cntv = 0;
    for (; u % i == 0; u /= i) cntu++;
    for (; v % i == 0; v /= i) cntv++;
    if (cntu || cntv) {
      int dif = abs(cntu - cntv);
      facinv = facinv * inv[dif] % mod;
      sum += dif;
    }
  }
  return fac[sum] * facinv % mod;
}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
inline void work() {
  long long D = read();
  divide(D);
  for (int i = 1; i <= 1000; i++) fac[i] = fac[i - 1] * i % mod;
  inv[1000] = q_pow(fac[1000], mod - 2);
  for (int i = 999; i; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  for (int q = read(); q; q--) {
    long long u = read(), v = read(), g = gcd(u, v);
    printf("%lld\n", count(u, g) * count(v, g) % mod);
  }
}
int main() {
  work();
  return 0;
}

```