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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
std::pair<int, int> DR[] = {{-1, 0}, {0, 1},   {1, 0}, {0, -1},
                            {-1, 1}, {-1, -1}, {1, 1}, {1, -1}};
using namespace std;
const long long mod = 998244353;
long long nop = 100;
vector<int> primes(nop, 1);
void sieve() {
  primes[1] = primes[0] = 0;
  for (long long i = 2; i * i <= nop; i++) {
    if (primes[i]) {
      for (long long j = i * i; j <= nop; j += i) {
        primes[j] = 0;
      }
    }
  }
}
long long gcd(long long a, long long b) {
  if (b > a) swap(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
void extgcd(long long a, long long b, long long& x, long long& y) {
  if (a == 0) {
    x = a;
    y = b;
    return;
  }
  extgcd(b % a, a, x, y);
  long long y1 = y, x1 = x;
  x = y1 - (b / a) * x1;
  y = x1;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long bexp(long long num, long long e) {
  long long ans = 1;
  while (e > 0) {
    if (e & 1) {
      ans *= num;
    }
    num *= num;
    e >>= 1;
  }
  return ans;
}
long long mexp(long long num, long long e) {
  long long ans = 1;
  while (e > 0) {
    if (e & 1) {
      ans = (ans * num) % mod;
    }
    num = (num * num) % mod;
    e >>= 1;
  }
  return ans % mod;
}
long long modinv(long long a) { return mexp(a, mod - 2); }
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
class PnC {
 public:
  vector<long long> fact;
  long long MAXN;
  PnC(long long MAX) {
    MAXN = MAX;
    fact.resize(MAXN + 1, 1);
    for (long long i = 2; i <= (MAXN); i++) {
      fact[i] = (fact[i - 1] * i) % mod;
    }
  }
  long long P(long long n, long long r) {
    return (fact[n] * modinv(fact[n - r])) % mod;
  }
  long long C(long long n, long long r) {
    return (fact[n] * (modinv(fact[r]) * modinv(fact[n - r])) % mod) % mod;
  }
};
PnC pp(70LL);
long long calc(long long u, vector<long long>& primes) {
  long long cnt = 0, ans = 1;
  for (long long i = 0; i < primes.size() && primes[i] * primes[i] <= u; i++) {
    long long tmp = 0;
    while (u % primes[i] == 0) {
      cnt++;
      u /= primes[i];
      tmp++;
    }
    if (tmp > 1) ans = (ans * modinv(pp.fact[tmp])) % mod;
  }
  if (u > 1) {
    cnt++;
  }
  ans = (ans * pp.fact[cnt]) % mod;
  return ans;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  int t = 1;
  for (long long T = 1; T <= (t); T++) {
    long long D;
    cin >> D;
    long long tmp = D;
    vector<long long> primes;
    for (long long i = 2; i * i <= tmp; i++) {
      if (tmp % i == 0) {
        primes.push_back(i);
      }
      while (tmp % i == 0) {
        tmp /= i;
      }
    }
    if (tmp > 1) {
      primes.push_back(tmp);
    }
    long long q;
    cin >> q;
    for (long long i = 0; i < (q); i++) {
      long long u, v;
      cin >> u >> v;
      long long g = gcd(u, v);
      u /= g;
      v /= g;
      long long ans = 1;
      ans = (ans * calc(u, primes)) % mod;
      ans = (ans * calc(v, primes)) % mod;
      cout << ans << endl;
    }
  }
  return 0;
}