#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  bool f = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x = f ? -x : x;
}
const long long N = 3e5;
const long long mod = 998244353;
long long n;
long long d, q;
long long p[20];
long long fac[60];
long long inv[60];
inline long long qw(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long now[20];
long long nn = 1;
inline void pre() {
  inv[0] = inv[1] = fac[0] = fac[1] = 1;
  for (long long i = (2); i <= (52); ++i) fac[i] = (i * fac[i - 1]) % mod;
  inv[52] = qw(fac[52], mod - 2);
  for (long long i = 51; i >= 2; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
  long long h = d;
  long long hhh = sqrt(d);
  for (long long i = (2); i <= (hhh); ++i) {
    if (!(h % i)) {
      p[++n] = i;
      while (!(h % i)) h /= i;
    }
  }
  if (h > 1) p[++n] = h;
}
inline long long query(long long x) {
  long long sum = 0;
  long long ans = 1;
  for (long long i = 1; i <= n; ++i) {
    long long siz = 0;
    while (!(x % p[i])) ++siz, x /= p[i];
    ans = (ans * inv[siz]) % mod;
    sum += siz;
  }
  ans = (ans * fac[sum]) % mod;
  return ans;
}
long long gcd(long long u, long long v) { return v ? gcd(v, u % v) : u; }
int main() {
  read(d);
  read(q);
  pre();
  while (q--) {
    long long u, v;
    read(u);
    read(v);
    if (u < v) swap(u, v);
    long long w = gcd(u, v);
    long long ans = query(u / w) * query(v / w) % mod;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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>
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  bool f = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x = f ? -x : x;
}
const long long N = 3e5;
const long long mod = 998244353;
long long n;
long long d, q;
long long p[20];
long long fac[60];
long long inv[60];
inline long long qw(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long now[20];
long long nn = 1;
inline void pre() {
  inv[0] = inv[1] = fac[0] = fac[1] = 1;
  for (long long i = (2); i <= (52); ++i) fac[i] = (i * fac[i - 1]) % mod;
  inv[52] = qw(fac[52], mod - 2);
  for (long long i = 51; i >= 2; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
  long long h = d;
  long long hhh = sqrt(d);
  for (long long i = (2); i <= (hhh); ++i) {
    if (!(h % i)) {
      p[++n] = i;
      while (!(h % i)) h /= i;
    }
  }
  if (h > 1) p[++n] = h;
}
inline long long query(long long x) {
  long long sum = 0;
  long long ans = 1;
  for (long long i = 1; i <= n; ++i) {
    long long siz = 0;
    while (!(x % p[i])) ++siz, x /= p[i];
    ans = (ans * inv[siz]) % mod;
    sum += siz;
  }
  ans = (ans * fac[sum]) % mod;
  return ans;
}
long long gcd(long long u, long long v) { return v ? gcd(v, u % v) : u; }
int main() {
  read(d);
  read(q);
  pre();
  while (q--) {
    long long u, v;
    read(u);
    read(v);
    if (u < v) swap(u, v);
    long long w = gcd(u, v);
    long long ans = query(u / w) * query(v / w) % mod;
    cout << ans << endl;
  }
  return 0;
}

```