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 JAVA):

Solution:
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
If I'm the sun, you're the moon
Because when I go up, you go down
*******************************
I'm working for the day I will surpass you
https://www.a2oj.com/Ladder16.html
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1334E
   {
      static long MOD = 998244353L;
      public static void main(String omkar[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         long D = Long.parseLong(st.nextToken());
         pdiv(D);
         int Q = Integer.parseInt(infile.readLine());
         StringBuilder sb = new StringBuilder();
         fac = new long[201];
         invfac = new long[201];
         fac[0] = 1L;   invfac[0] = 1L;
         for(int i=1; i <= 200; i++)
         {
            fac[i] = (fac[i-1]*i)%MOD;
            invfac[i] = power(fac[i], MOD-2, MOD);
         }
         while(Q-->0)
         {
            st = new StringTokenizer(infile.readLine());
            long X = Long.parseLong(st.nextToken());
            long Y = Long.parseLong(st.nextToken());
            long gcd = gcd(X,Y);
            //may or may not tle
            long res = (solve(X/gcd)*solve(Y/gcd))%MOD;
            sb.append(res+"\n");
         }
         System.out.print(sb);
      }
      static long[] fac;
      static long[] invfac;
      static ArrayList<Long> primes = new ArrayList<Long>();
      public static long solve(long X)
      {
         ArrayList<Integer> ls = new ArrayList<Integer>();
         for(long d: primes)
            if(X%d == 0)
            {
               int cnt = 0;
               while(X%d == 0)
               {
                  cnt++;
                  X /= d;
               }
               ls.add(cnt);
            }
         //combine
         int sum = 0;
         long res = 1L;
         for(int x: ls)
         {
            res = (res*invfac[x])%MOD;
            sum += x;
         }
         res = (res*fac[sum])%MOD;
         return res;
      }
      public static void pdiv(long D)
      {
         if(D%2 == 0)
         {
            primes.add(2L);
            while(D%2 == 0)
               D /= 2;
         }
         for(long d=3; d <= (long)(Math.sqrt(D)+0.0001); d+=2)
            if(D%d == 0)
            {
               while(D%d == 0)
                  D /= d;
               primes.add(d);
            }
         if(D > 2)
            primes.add(D);
      }
      public static long gcd(long a, long b)
      {
         if(a > b)
         {
            long t = a;
            a = b;
            b = t;
         }
         if(a == 0L)
            return b;
         return gcd(b%a, a);
      }
      public static long power(long x, long y, long p) 
      { 
          long res = 1L;      
          x = x % p;  
          while (y > 0) 
          { 
              if((y & 1)==1) 
                  res = (res * x) % p; 
              y = y >> 1;  
              x = (x * x) % p;  
          } 
          return res; 
      }
   }