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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.util.HashMap;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskE {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            long D = in.nextLong();
            int q = in.nextInt();
            ArrayList<Long> primes = new ArrayList<>();
            HashMap<Long, Integer> map = new HashMap<>();
            for (long i = 1; i <= Math.sqrt(D); i++) {
                if (D % i == 0) {
                    map.putIfAbsent(i, map.size());
                    map.putIfAbsent(D / i, map.size());
                }
            }
            for (long i = 2; i <= Math.sqrt(D); i++) {
                if (D % i == 0) {
                    primes.add(i);
                    while (D % i == 0) {
                        D /= i;
                    }
                }
            }
            if (D > 1) {
                primes.add(D);
            }
            int[][] primeArr = new int[map.size()][primes.size()];
            for (long val : map.keySet()) {
                long tempVal = val;
                int i = map.get(val);
                for (int j = 0; j < primes.size(); j++) {
                    while (tempVal % primes.get(j) == 0) {
                        tempVal /= primes.get(j);
                        primeArr[i][j]++;
                    }
                }
            }
            long[] factorial = new long[1000];
            long[] invFactorial = new long[1000];
            factorial[0] = 1;
            invFactorial[0] = 1;
            long MOD = 998244353;
            for (int i = 1; i < 1000; i++) {
                factorial[i] = factorial[i - 1] * i;
                factorial[i] %= MOD;
                invFactorial[i] = MathUtils.modFraction(1, factorial[i], MOD);
            }
            while (q-- > 0) {
                long u = in.nextLong();
                long v = in.nextLong();
                int[] first = primeArr[map.get(u)];
                int[] second = primeArr[map.get(v)];
                long res = 1;
                int a = 0;
                int b = 0;
                for (int i = 0; i < primes.size(); i++) {
                    if (first[i] < second[i]) {
                        a += Math.abs(first[i] - second[i]);
                    } else {
                        b += Math.abs(first[i] - second[i]);
                    }
                    res = (res * invFactorial[Math.abs(first[i] - second[i])]) % MOD;
                }
                res = ((res * factorial[a]) % MOD * factorial[b]) % MOD;
                out.println(res);
            }
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

    }

    static class MathUtils {
        public static long inverse(long a, long mod) {
            long[] inv = extended_gcd(a, mod);
            if (inv[0] != 1) {
                return 0;
            } else {
                return (inv[1] + 2 * mod) % mod;
            }
        }

        public static long modFraction(long a, long b, long mod) {
            //a is numerator, b is denominator
            a %= mod;
            b %= mod;
            long invB = inverse(b, mod);
            if (invB == 0) {
                int[] arr = new int[3];
                System.out.println(arr[4]);
                return -1;
            } else {
                return (invB * a) % mod;
            }
        }

        public static long[] extended_gcd(long a, long b) {
            //three numbers, first is gcd, second is x, third is y
            if (a == 0) {
                return new long[]{b, 0, 1};
            }
            long[] next = extended_gcd(b % a, a);
            long tempX = next[1];
            next[1] = next[2] - (b / a) * next[1];
            next[2] = tempX;
            return next;
        }

    }
}