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;
template <class Temp>
vector<Temp> read(long long n) {
  vector<Temp> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  return v;
}
template <class Temp>
void print(vector<long long> &v) {
  for (Temp el : v) {
    cout << el << " ";
  }
  cout << endl;
}
template <class Temp>
vector<vector<Temp>> matrix(long long n, long long m, long long value) {
  return vector<vector<Temp>>(n, vector<Temp>(m));
}
template <class Temp>
vector<vector<Temp>> readMatrix(long long n, long long m) {
  vector<vector<Temp>> matrix(n, vector<long long>(m, 0));
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      cin >> matrix[i][j];
    }
  }
  return matrix;
}
class CompressedPrefix {
  vector<long long> *array;
  vector<int> prefix;
  long long compressionRate;
  long long value;

 public:
  CompressedPrefix(vector<long long> *a, long long v, long long root) {
    array = a;
    value = v;
    compressionRate = root;
    prefix.resize(array->size() / compressionRate + 1);
    for (long long i = 0; i < array->size(); i++) {
      prefix[i / compressionRate] += (*array)[i] == value;
    }
    for (long long i = 1; i < prefix.size(); i++) {
      prefix[i] += prefix[i - 1];
    }
  }
  long long operator[](long long index) {
    long long blocks = index / compressionRate;
    long long sum;
    if (blocks > 0) {
      sum = prefix[blocks - 1];
    } else {
      sum = 0;
    }
    for (long long i = blocks * compressionRate; i <= index; i++) {
      sum += (*array)[i] == value;
    }
    return sum;
  }
};
vector<long long> primes;
long long mod = 998244353;
vector<long long> fact(100, 1);
long long power(long long n, long long p) {
  if (p == 0) {
    return 1;
  }
  if (p == 1) {
    return n;
  }
  long long ans = power(n, p / 2);
  ans *= ans;
  ans %= mod;
  if (p % 2 == 1) {
    ans *= n;
    ans %= mod;
  }
  return ans;
}
long long inverse(long long n) { return power(n, mod - 2); }
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
void breakUp(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      primes.push_back(i);
      n /= i;
      while (n % i == 0) {
        n /= i;
      }
    }
  }
  if (n != 1) {
    primes.push_back(n);
  }
}
void calcFactorial() {
  mod = 998244353;
  fact.resize(100, 1);
  for (long long i = 2; i < 100; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= mod;
  }
}
long long calcPaths(long long n) {
  long long sum = 0;
  long long devider = 1;
  for (long long prime : primes) {
    if (n % prime == 0) {
      long long count = 0;
      while (n % prime == 0) {
        n /= prime;
        count++;
      }
      sum += count;
      devider *= fact[count];
      devider %= mod;
    }
  }
  long long ans = fact[sum];
  ans *= inverse(devider);
  ans %= mod;
  return ans;
}
void solve(long long t) {
  long long d;
  cin >> d;
  breakUp(d);
  calcFactorial();
  long long q;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long u, v;
    cin >> u >> v;
    if (u < v) {
      swap(u, v);
    }
    long long g = gcd(u, v);
    u /= g;
    v /= g;
    long long ans = calcPaths(u) * calcPaths(v);
    ans %= mod;
    cout << ans << "\n";
  }
}
void test() {
  long long T;
  cin >> T;
  for (long long t = 1; t <= T; t++) {
    solve(0);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(8);
  solve(0);
}