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). 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
class modint {
 public:
  modint() : n_(0) {}
  modint(int n) : n_(n) {}
  int Int() { return n_; }
  modint operator+(modint y) { return add(n_, y.n_); }
  modint operator+=(modint y) {
    n_ = add(n_, y.n_);
    return *this;
  }
  modint operator-(modint y) { return add(n_, -y.n_); }
  modint operator-=(modint y) {
    n_ = add(n_, -y.n_);
    return *this;
  }
  modint operator*(modint y) { return mul(n_, y.n_); }
  modint operator*=(modint y) {
    n_ = mul(n_, y.n_);
    return *this;
  }
  modint operator/(modint y) { return divide(n_, y.n_); }
  modint operator/=(modint y) {
    n_ = divide(n_, y.n_);
    return *this;
  }
  modint operator^(long long y) { return binpow(n_, y); }
  modint operator^=(long long y) {
    n_ = binpow(n_, y);
    return *this;
  }

 private:
  int add(int x, int y) {
    x += y;
    while (x >= mod_) x -= mod_;
    while (x < 0) x += mod_;
    return x;
  }
  int mul(int x, int y) { return (x * 1ll * y) % mod_; }
  int binpow(int x, long long y) {
    int z = 1;
    while (y) {
      if (y & 1) z = mul(z, x);
      x = mul(x, x);
      y >>= 1;
    }
    return z;
  }
  int inv(int x) { return binpow(x, mod_ - 2); }
  int divide(int x, int y) { return mul(x, inv(y)); }
  static const int mod_ = 998244353;
  int n_;
};
long long gcd(long long x, long long y) {
  if (y == 0) {
    return x;
  }
  return gcd(y, x % y);
}
void Solve() {
  long long dd;
  int q;
  cin >> dd >> q;
  auto d = dd;
  vector<long long> fac;
  for (long long i = 2; i * i <= d; i++) {
    if (d % i == 0) {
      fac.push_back(i);
      while (d % i == 0) {
        d /= i;
      }
    }
  }
  if (d > 1) {
    fac.push_back(d);
  }
  map<long long, modint> init;
  init[1] = 1;
  function<modint(long long)> dfs;
  dfs = [&](long long num) {
    if (init.find(num) != end(init)) {
      return init[num];
    }
    auto& count = init[num];
    for (auto f : fac) {
      if (num % f == 0) {
        count += dfs(num / f);
      }
    }
    return count;
  };
  dfs(dd);
  long long u, v;
  for (int i = 0; i < q; i++) {
    cin >> u >> v;
    modint ans = 1;
    auto c = gcd(u, v);
    ans *= init[u / c];
    ans *= init[v / c];
    printf("%d\n", ans.Int());
  }
}
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t = 1;
  while (t--) {
    Solve();
  }
  return 0;
}