#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
const long long mod = 998244353;
const int maxn = (1e5) + 10;
long long n, q, x, y, prime[maxn], jc[maxn], ivjc[maxn], iv[maxn], ans;
int tot, cnt, s1, s2;
vector<int> a, b;
int main() {
  read(n);
  for (long long i = 2; i * i <= n; i++) {
    if (n % i != 0) continue;
    prime[++tot] = i;
    while (n % i == 0) n /= i;
  }
  if (n > 1) prime[++tot] = n;
  jc[0] = iv[0] = jc[1] = iv[1] = ivjc[0] = ivjc[1] = 1;
  for (int i = 2; i < maxn; i++) {
    jc[i] = jc[i - 1] * i % mod;
    iv[i] = (mod - mod / i) * iv[mod % i] % mod;
    ivjc[i] = ivjc[i - 1] * iv[i] % mod;
  }
  read(q);
  while (q--) {
    read(x);
    read(y);
    a.clear();
    b.clear();
    s1 = s2 = 0;
    for (int i = 1; i <= tot; i++) {
      cnt = 0;
      while (x % prime[i] == 0) x /= prime[i], cnt++;
      while (y % prime[i] == 0) y /= prime[i], cnt--;
      if (cnt < 0)
        a.push_back(-cnt), s1 -= cnt;
      else
        b.push_back(cnt), s2 += cnt;
    }
    ans = jc[s1] * jc[s2] % mod;
    for (int i = 0; i < a.size(); i++) ans = ans * ivjc[a[i]] % mod;
    for (int i = 0; i < b.size(); i++) ans = ans * ivjc[b[i]] % mod;
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
using namespace std;
template <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
const long long mod = 998244353;
const int maxn = (1e5) + 10;
long long n, q, x, y, prime[maxn], jc[maxn], ivjc[maxn], iv[maxn], ans;
int tot, cnt, s1, s2;
vector<int> a, b;
int main() {
  read(n);
  for (long long i = 2; i * i <= n; i++) {
    if (n % i != 0) continue;
    prime[++tot] = i;
    while (n % i == 0) n /= i;
  }
  if (n > 1) prime[++tot] = n;
  jc[0] = iv[0] = jc[1] = iv[1] = ivjc[0] = ivjc[1] = 1;
  for (int i = 2; i < maxn; i++) {
    jc[i] = jc[i - 1] * i % mod;
    iv[i] = (mod - mod / i) * iv[mod % i] % mod;
    ivjc[i] = ivjc[i - 1] * iv[i] % mod;
  }
  read(q);
  while (q--) {
    read(x);
    read(y);
    a.clear();
    b.clear();
    s1 = s2 = 0;
    for (int i = 1; i <= tot; i++) {
      cnt = 0;
      while (x % prime[i] == 0) x /= prime[i], cnt++;
      while (y % prime[i] == 0) y /= prime[i], cnt--;
      if (cnt < 0)
        a.push_back(-cnt), s1 -= cnt;
      else
        b.push_back(cnt), s2 += cnt;
    }
    ans = jc[s1] * jc[s2] % mod;
    for (int i = 0; i < a.size(); i++) ans = ans * ivjc[a[i]] % mod;
    for (int i = 0; i < b.size(); i++) ans = ans * ivjc[b[i]] % mod;
    printf("%lld\n", ans);
  }
  return 0;
}

```