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;
long long d, s, t;
int q;
const int mod = 998244353;
long long gcd(long long x, long long y) {
  if (!y) return x;
  return gcd(y, x % y);
}
int zs[10000007];
bool b[200000007];
int num[1007];
long long d2;
int ys[1007];
long long jc[107];
long long X, Y;
int js2 = 0;
inline void exgcd(long long A, long long B) {
  if (!B) {
    X = 1;
    Y = 0;
    return;
  }
  exgcd(B, A % B);
  long long t = X;
  X = Y;
  Y = t - A / B * Y;
}
inline long long find(long long a) {
  if (a == 1) return 1;
  long long fz = 0, fm = 1;
  if (d2 != 1) {
    if (!(a % d2)) ++fz;
  }
  for (register int i = 1; i <= js2; ++i) {
    long long ei = 0;
    while (!(a % ys[i])) {
      ++ei;
      a /= ys[i];
    }
    if (ei) {
      fz += ei;
      fm = fm * jc[ei] % mod;
    }
  }
  fz = jc[fz];
  exgcd(fm, mod);
  return fz * ((X % mod + mod) % mod) % mod;
}
int main() {
  cin >> d >> q;
  jc[1] = 1;
  for (register int i = 2; i <= 64; ++i) jc[i] = jc[i - 1] * i % mod;
  int js = 0;
  int D = sqrt(d);
  for (register int i = 2; i <= D; ++i) {
    if (!b[i]) {
      ++js;
      zs[js] = i;
    }
    for (register int j = 1; j <= js && i * zs[j] <= D; ++j) {
      b[i * zs[j]] = 1;
      if (!(i % zs[j])) break;
    }
  }
  d2 = d;
  for (register int i = 1; i <= js; ++i) {
    if (!(d2 % zs[i])) {
      ++js2;
      ys[js2] = zs[i];
    }
    while (!(d2 % zs[i])) {
      d2 /= zs[i];
      ++num[i];
    }
  }
  if (d2 > 1) ++num[0];
  for (register int i = 1; i <= q; ++i) {
    cin >> s >> t;
    long long g = gcd(s, t);
    s = s / g;
    t = t / g;
    cout << find(s) * find(t) % mod << endl;
    ;
  }
  return 0;
}