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:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.HashSet;
import java.util.StringTokenizer;

public class divisorpaths {
	public static void main(String[] args) {
		FastScanner scan=new FastScanner();
		PrintWriter out=new PrintWriter(System.out);
		
		/*
		we must start at A
		divide repeatedly by one prime factor
		then multiply to get back up to B.
		
		all common factors to A and B must stay
		A-->gcd(A,B)-->B
		
		we need to go down one factor at a time because otherwise there would not be
		an edge between them (wouldn't fulfill the x/y is prime req)
		each of these paths is the shortest because we are taking out one
		factor at a time and it doesn't matter what it is
		
		we have to count the number of ways to order the ones we take
		we use the multinomial coefficient
		*/
		
		long n=scan.nextLong();
		int t=scan.nextInt();
		
		nfactors=new HashSet<>();
		HashMap<Long,Integer> map=factorize(n);
		fac=new long[51];
		invfac=new long[51];
		fac[0]=invfac[0]=1;
		
		for(int i=1;i<fac.length;i++) {
			fac[i]=mult(fac[i-1],i);
			invfac[i]=pow(fac[i],MOD-2);
		}
		for(int tt=0;tt<t;tt++) {
			long A=scan.nextLong(), B=scan.nextLong();
			long I=gcd(A,B);
			
			out.println(mult(calc(A,I),calc(B,I)));
		}
		out.close();
	}
	static long[] fac,invfac;
	public static long calc(long A, long I) {
		long p1=1L;
		int total=0;
		for(long f:nfactors) {
			long x=1L;
			int delta=0;
			while(A%x==0) {
				if(A%x==0&&I%x!=0) delta++;
				x*=f;
			}
//			System.out.println(f+" "+delta+" "+invfac[delta]);
			p1=mult(p1,invfac[delta]);
			total+=delta;
		}
		p1=mult(p1,fac[total]);
		return p1;
	}
	static final long MOD=998244353;
	public static long mult(long a, long b) {
		return ((a%MOD)*(b%MOD))%MOD;
	}
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			try	{
				br = new BufferedReader(new InputStreamReader(System.in));
				st = new StringTokenizer(br.readLine());
			} catch (Exception e){e.printStackTrace();}
		}

		public String next() {
			if (st.hasMoreTokens())	return st.nextToken();
			try {st = new StringTokenizer(br.readLine());}
			catch (Exception e) {e.printStackTrace();}
			return st.nextToken();
		}

		public int nextInt() {return Integer.parseInt(next());}

		public long nextLong() {return Long.parseLong(next());}

		public double nextDouble() {return Double.parseDouble(next());}

		public String nextLine() {
			String line = "";
			if(st.hasMoreTokens()) line = st.nextToken();
			else try {return br.readLine();}catch(IOException e){e.printStackTrace();}
			while(st.hasMoreTokens()) line += " "+st.nextToken();
			return line;
		}
	}
	public static long pow(long b, long e) {
		if(e==0L) return 1L;
		if(e%2==0) {
			long temp=pow(b,e/2);
			return mult(temp,temp);
		}
		return mult(b,pow(b,e-1));
	}
	static HashSet<Long> nfactors;
	public static HashMap<Long,Integer> factorize(long x) {
		HashMap<Long,Integer> f=new HashMap<>();
		for(long i=2;i*i<=x;i++) {
			int k=0;
			while(x%i==0) {
				x/=i;
				k++;
			}
			if(k>0) {
				f.put(i,k);
				nfactors.add(i);
			}
		}
		if(x>1) nfactors.add(x);
		return f;
	}
	public static long gcd(long a, long b) {
		if(b==0) return a;
		return gcd(b,a%b);
	}
}