#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  long long c;
  while (b != 0) {
    c = a % b;
    a = b;
    b = c;
  }
  return a;
}
bool je(long long int a) {
  if (a == 1) return false;
  if (a == 2) return true;
  if (a == 79) return true;
  long long int i = 2;
  while (i * i <= a) {
    if (a % i == 0) return false;
    i++;
  }
  return true;
}
vector<long long int> rozloz(long long int a) {
  vector<long long int> vys;
  const long long int aa = a;
  bool br = false;
  if (!je(a))
    for (long long int i = 2; i * i <= aa; i++) {
      while (a % i == 0) {
        a /= i;
        if (vys.empty() || vys.back() != i) vys.push_back(i);
        if (je(a)) {
          if (vys.empty() || vys.back() != a) vys.push_back(a);
          br = true;
          break;
        }
      }
      if (br) break;
    }
  return vys;
}
vector<unsigned long long> f(100, 1), p(100, 0);
unsigned long long moc(unsigned long long a, unsigned long long b,
                       unsigned long long c) {
  if (b == 0) return 1;
  unsigned long long pol = moc(a, b / 2, c);
  pol *= pol;
  pol %= c;
  if (b % 2 == 1) {
    pol *= a;
  }
  return pol % c;
}
unsigned long long invp(unsigned long long a) {
  return moc(a, 998244353 - 2, 998244353);
}
unsigned long long n_nad_k(long long n, long long k) {
  if (n < k) return 0;
  if (n / 2 < k) k = n - k;
  if (k == 0) return 1;
  if (k < 0) return 0;
  if (998244353 <= n) {
    unsigned long long dol = invp(f[k]) * invp(f[n - k]);
    unsigned long long ci = dol, cp = 998244353, cn = 0;
    while (cp < n) {
      cn += k / cp + ((n - k) / cp);
      cp *= 998244353;
    }
    if (cn != p[n]) return 0;
  }
  unsigned long long vys =
      ((((f[n] * invp(f[k])) % 998244353) * invp(f[n - k])) % 998244353) %
      998244353;
  return vys;
}
vector<vector<long long> > c(100, vector<long long>(100));
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (unsigned long long i = 1; i < 100; i++) {
    p[i] = p[i - 1];
    unsigned long long ci = i;
    unsigned long long cp = 998244353;
    while (ci % cp == 0) {
      cp *= 998244353;
      p[i]++;
    }
    f[i] = (f[i - 1] * (i / (cp / 998244353))) % 998244353;
  }
  for (unsigned long long i = 0; i < 100; i++) {
    for (unsigned long long j = 0; j < 100; j++) {
      c[i][j] = n_nad_k(i, j);
    }
  }
  long long d, q;
  cin >> d >> q;
  vector<long long> pri = rozloz(d);
  while (q--) {
    long long v, u;
    cin >> v >> u;
    long long g = gcd(v, u);
    long long vg = v / g, ug = u / g;
    long long cu = 1, cv = 1, su = 0, sv = 0;
    for (long long i = 0; i < pri.size(); i++) {
      long long cn = 0;
      while (vg % pri[i] == 0) {
        vg /= pri[i];
        cn++;
      }
      cu *= c[su + cn][cn];
      cu %= 998244353;
      su += cn;
      cn = 0;
      while (ug % pri[i] == 0) {
        ug /= pri[i];
        cn++;
      }
      cv *= c[sv + cn][cn];
      cv %= 998244353;
      sv += cn;
    }
    cout << (cu * cv) % 998244353 << "\n";
  }
  return 0;
}

### Prompt

Generate a CPP solution to 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;
long long gcd(long long a, long long b) {
  long long c;
  while (b != 0) {
    c = a % b;
    a = b;
    b = c;
  }
  return a;
}
bool je(long long int a) {
  if (a == 1) return false;
  if (a == 2) return true;
  if (a == 79) return true;
  long long int i = 2;
  while (i * i <= a) {
    if (a % i == 0) return false;
    i++;
  }
  return true;
}
vector<long long int> rozloz(long long int a) {
  vector<long long int> vys;
  const long long int aa = a;
  bool br = false;
  if (!je(a))
    for (long long int i = 2; i * i <= aa; i++) {
      while (a % i == 0) {
        a /= i;
        if (vys.empty() || vys.back() != i) vys.push_back(i);
        if (je(a)) {
          if (vys.empty() || vys.back() != a) vys.push_back(a);
          br = true;
          break;
        }
      }
      if (br) break;
    }
  return vys;
}
vector<unsigned long long> f(100, 1), p(100, 0);
unsigned long long moc(unsigned long long a, unsigned long long b,
                       unsigned long long c) {
  if (b == 0) return 1;
  unsigned long long pol = moc(a, b / 2, c);
  pol *= pol;
  pol %= c;
  if (b % 2 == 1) {
    pol *= a;
  }
  return pol % c;
}
unsigned long long invp(unsigned long long a) {
  return moc(a, 998244353 - 2, 998244353);
}
unsigned long long n_nad_k(long long n, long long k) {
  if (n < k) return 0;
  if (n / 2 < k) k = n - k;
  if (k == 0) return 1;
  if (k < 0) return 0;
  if (998244353 <= n) {
    unsigned long long dol = invp(f[k]) * invp(f[n - k]);
    unsigned long long ci = dol, cp = 998244353, cn = 0;
    while (cp < n) {
      cn += k / cp + ((n - k) / cp);
      cp *= 998244353;
    }
    if (cn != p[n]) return 0;
  }
  unsigned long long vys =
      ((((f[n] * invp(f[k])) % 998244353) * invp(f[n - k])) % 998244353) %
      998244353;
  return vys;
}
vector<vector<long long> > c(100, vector<long long>(100));
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (unsigned long long i = 1; i < 100; i++) {
    p[i] = p[i - 1];
    unsigned long long ci = i;
    unsigned long long cp = 998244353;
    while (ci % cp == 0) {
      cp *= 998244353;
      p[i]++;
    }
    f[i] = (f[i - 1] * (i / (cp / 998244353))) % 998244353;
  }
  for (unsigned long long i = 0; i < 100; i++) {
    for (unsigned long long j = 0; j < 100; j++) {
      c[i][j] = n_nad_k(i, j);
    }
  }
  long long d, q;
  cin >> d >> q;
  vector<long long> pri = rozloz(d);
  while (q--) {
    long long v, u;
    cin >> v >> u;
    long long g = gcd(v, u);
    long long vg = v / g, ug = u / g;
    long long cu = 1, cv = 1, su = 0, sv = 0;
    for (long long i = 0; i < pri.size(); i++) {
      long long cn = 0;
      while (vg % pri[i] == 0) {
        vg /= pri[i];
        cn++;
      }
      cu *= c[su + cn][cn];
      cu %= 998244353;
      su += cn;
      cn = 0;
      while (ug % pri[i] == 0) {
        ug /= pri[i];
        cn++;
      }
      cv *= c[sv + cn][cn];
      cv %= 998244353;
      sv += cn;
    }
    cout << (cu * cv) % 998244353 << "\n";
  }
  return 0;
}

```