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;
const long long mod = 998244353;
map<long long, vector<pair<long long, int>>> primeDiv;
vector<long long> fact;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void fillFact() {
  fact.push_back(1);
  for (int i = 1; i <= 100; i++) fact.push_back((fact.back() * i) % mod);
}
void getDivisorsHelper(int i, long long curr, vector<pair<long long, int>> &p,
                       vector<pair<long long, int>> &v) {
  if (i == (int)v.size()) {
    primeDiv[curr] = p;
    return;
  }
  getDivisorsHelper(i + 1, curr, p, v);
  p.push_back({v[i].first, 0});
  for (int j = 0; j < v[i].second; j++) {
    curr *= v[i].first;
    p.back().second++;
    getDivisorsHelper(i + 1, curr, p, v);
  }
  p.pop_back();
}
void getDivisors(long long x) {
  vector<pair<long long, int>> v;
  vector<pair<long long, int>> p;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i) {
      continue;
    }
    int c = 0;
    while (!(x % i)) {
      c++;
      x /= i;
    }
    v.push_back({i, c});
  }
  if (x != 1) {
    v.push_back({x, 1});
  }
  getDivisorsHelper(0, 1, p, v);
}
long long power(long long x, long long n) {
  if (!n) {
    return 1;
  }
  long long res = power(x, n >> 1);
  res = (res * res) % mod;
  if (n & 1) {
    return (res * x) % mod;
  }
  return res;
}
long long modInv(long long x) { return power(x, mod - 2); }
long long findPath(long long s, long long d) {
  if (s == d) {
    return 1;
  }
  long long target = s / d;
  long long num = 0, denum = 1;
  for (auto &x : primeDiv[target]) {
    denum *= fact[x.second];
    num += x.second;
    denum %= mod;
  }
  num = fact[(int)num];
  return (num * modInv(denum)) % mod;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  long long n;
  int q;
  cin >> n >> q;
  getDivisors(n);
  fillFact();
  while (q--) {
    long long u, v;
    cin >> u >> v;
    long long g = gcd(u, v);
    cout << (findPath(u, g) * findPath(v, g)) % mod << '\n';
  }
  cin.ignore(2);
  return 0;
}