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;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int ddx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, ddy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
void mad(long long &a, long long b) {
  a = (a + b) % 998244353;
  if (a < 0) a += 998244353;
}
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
vector<long long> prime;
bool isnotp[40000000];
void sieve(long long n) {
  for (long long i = 2; i <= n; i++) {
    if (isnotp[i]) continue;
    for (long long j = i + i; j <= n; j += i) {
      isnotp[j] = 1;
    }
    prime.push_back(i);
  }
}
long long powmod(long long a, long long b) {
  if (b == 0) return 1;
  long long c = powmod(a, b / 2ll);
  c = ((c * c) % 998244353 * ((b & 1ll) ? a : 1ll)) % 998244353;
  return c;
}
map<long long, map<long long, int> > mp;
void solve() {
  long long n;
  cin >> n;
  long long nn = sqrt(n);
  sieve(nn);
  vector<long long> factors, pf;
  for (long long i = 1; i <= nn; i++) {
    if (n % i) continue;
    factors.push_back(i);
    if (i * i != n) factors.push_back(n / i);
  }
  long long m = n;
  for (long long i : prime) {
    if (i * i > m) break;
    if (m % i == 0) pf.push_back(i);
    while (m % i == 0) {
      m /= i;
    }
  }
  if (m > 1) pf.push_back(m);
  sort(factors.begin(), factors.end());
  sort(pf.begin(), pf.end());
  int q;
  cin >> q;
  for (long long i : factors) {
    long long f = i;
    for (long long j : pf) {
      int cnt = 0;
      while (f % j == 0) {
        f /= j;
        cnt++;
      }
      mp[i][j] = cnt;
    }
  }
  long long fact[100], ifact[100];
  fact[0] = 1;
  ifact[0] = 1;
  for (long long i = 1; i < 100; i++) {
    fact[i] = (fact[i - 1] * i) % 998244353;
    ifact[i] = (powmod(fact[i], 998244353 - 2));
  }
  while (q--) {
    long long u, v;
    cin >> u >> v;
    long long ans1 = 1, ans2 = 1, ans = 1;
    long long sum1 = 0, sum2 = 0;
    long long g = gcd(u, v);
    for (long long i : pf) {
      long long dif = max(0, (mp[v][i] - mp[g][i]));
      sum1 += dif;
      ans1 = (ans1 * ifact[dif]) % 998244353;
    }
    for (long long i : pf) {
      long long dif = max(0, (mp[u][i] - mp[g][i]));
      sum2 += dif;
      ans2 = (ans2 * ifact[dif]) % 998244353;
    }
    ans1 = (ans1 * fact[sum1]) % 998244353;
    ans2 = (ans2 * fact[sum2]) % 998244353;
    ans = (ans1 * ans2) % 998244353;
    cout << ans << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}