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,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
void err(istream_iterator<string> it) { cout << '\n'; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << "[" << *it << " = " << a << "] ";
  err(++it, args...);
}
pair<int, int> dx[] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
const int inf = 2e9 + 7;
const long long INF = 3e18 + 7;
const long double pi = 3.1415926535897932384626433;
const double eps = 1e-6;
const long double EPS = 1e-9;
long long mod = 1e9 + 7;
const long long MOD = 1e15 + 7;
const bool multiple_tests = 0;
const int N = 1e6 + 7;
mt19937 lol(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rofl(chrono::steady_clock::now().time_since_epoch().count());
long long bp(long long a, long long n) {
  if (n == 0) return 1;
  long long res = bp(a, n / 2);
  res = res * res % mod;
  if (n % 2) res = res * a % mod;
  return res;
}
long long inv(long long a) { return bp(a, mod - 2); }
long long fact[101], rfact[101];
void solve() {
  mod = 998244353;
  long long d;
  cin >> d;
  long long sq = sqrt(d);
  vector<long long> p;
  for (long long i = 2; i <= sq; i++) {
    if (d % i == 0) {
      p.push_back(i);
      while (d % i == 0) d /= i;
    }
  }
  if (d > 1) p.push_back(d);
  fact[0] = 1;
  for (long long i = 1; i <= 100; i++) fact[i] = fact[i - 1] * i % mod;
  rfact[100] = inv(fact[100]);
  for (long long i = 99; i >= 0; i--) {
    rfact[i] = rfact[i + 1] * (i + 1) % mod;
  }
  long long q;
  cin >> q;
  while (q--) {
    long long a, b;
    long long suma = 0, sumb = 0;
    cin >> a >> b;
    vector<long long> down, up;
    for (long long e : p) {
      long long cnt = 0;
      while (a % e == 0) {
        cnt++;
        a /= e;
      }
      while (b % e == 0) {
        cnt--;
        b /= e;
      }
      if (cnt > 0)
        down.push_back(cnt), suma += cnt;
      else
        up.push_back(-cnt), sumb += -cnt;
    }
    long long ans = fact[suma];
    for (long long e : down) {
      ans = ans * rfact[e] % mod;
    }
    ans = ans * fact[sumb] % mod;
    for (long long e : up) {
      ans = ans * rfact[e] % mod;
    }
    cout << ans << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long TT = 1;
  if (multiple_tests) cin >> TT;
  for (long long testcase = 1; testcase <= TT; testcase++) {
    solve();
    cout << '\n';
  }
  return 0;
}