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.util.*;
import java.io.*;

public class E {
	static int MOD = 998244353;

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		long D = Long.parseLong(in.readLine());
		long Dtemp = D;
		int q = Integer.parseInt(in.readLine());
		long[] first = new long[q];
		long[] second = new long[q];
		for (int a = 0; a < q; a++) {
			StringTokenizer token = new StringTokenizer(in.readLine());
			first[a] = Long.parseLong(token.nextToken());
			second[a] = Long.parseLong(token.nextToken());
		}
		ArrayList<Long> primes = new ArrayList<Long>(55);
		for (long a = 2; a * a <= D; a++)
			if (D % a == 0) {
				primes.add(a);
				while (D % a == 0)
					D /= a;
			}
		if (D > 1)
			primes.add(D);
		D = Dtemp;
		long[] fact = new long[55];
		long[] invfact = new long[55];
		fact[0] = 1;
		for (int a = 1; a < 55; a++)
			fact[a] = (fact[a - 1] * a) % MOD;
		invfact[0] = 1;
		invfact[1] = 1;
		for (int a = 2; a < 55; a++)
			invfact[a] = exp(fact[a], MOD - 2);
		for (int a = 0; a < q; a++) {
			long GCD = gcd(first[a], second[a]);
			long f = first[a] / GCD;
			long s = second[a] / GCD;
			long answer = 1;
			int sum = 0;
			long product = 1;
			for (int b = 0; b < primes.size(); b++) {
				int count = 0;
				while(f % primes.get(b) == 0) {
					count++;
					f = f / primes.get(b);
				}
				sum += count;
				product = (product * invfact[count]) % MOD;
			}
			answer = (fact[sum] * product) % MOD;
			sum = 0;
			product = 1;
			for (int b = 0; b < primes.size(); b++) {
				int count = 0;
				while(s % primes.get(b) == 0) {
					count++;
					s = s / primes.get(b);
				}
				sum += count;
				product = (product * invfact[count]) % MOD;
			}
			answer = (answer * ((fact[sum] * product) % MOD)) % MOD;
			System.out.println(answer);
		}
	}

	static long gcd(long a, long b) {
		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static long exp(long base, int power) {
		if (power == 0)
			return 1;
		if (power == 1)
			return base;
		long ans = exp(base, power / 2);
		ans = (ans * ans) % MOD;
		if (power % 2 == 1)
			ans = (ans * base) % MOD;
		return ans;
	}
}