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;
const long long MAXI = 1234567890123456LL;
const long long SQRTMAXI = 32622776;
const long long MOD = 998244353LL;
const long long N = 111;
vector<bool> sieve;
vector<int> primes;
vector<int> factor;
unordered_map<long long, long long> mp;
void init(long long n) {
  sieve = vector<bool>(SQRTMAXI, false);
  for (long long i = 2; i < SQRTMAXI; i++) {
    if (sieve[i] == true) {
      continue;
    }
    for (long long j = 2; i * j < SQRTMAXI; j++) {
      sieve[i * j] = true;
    }
  }
  primes.push_back(2);
  for (int i = 3; i < SQRTMAXI; i += 2) {
    if (sieve[i] == false && n % i == 0) {
      primes.push_back(i);
    }
  }
  factor.push_back(1);
  for (int i = 1; i < N; i++) {
    factor.push_back(1LL * i * factor.back() % MOD);
  }
}
long long gcd(long long a, long long b) {
  if (a % b == 0) {
    return b;
  }
  return gcd(b, a % b);
}
long long mypow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a % MOD;
    }
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}
long long divmod(long long a) { return mypow(a, MOD - 2) % MOD; }
long long calcpath(long long u) {
  const long long key = u;
  if (mp.count(key)) {
    return mp[key];
  }
  long long tot = 0;
  long long div = 1;
  for (auto p : primes) {
    if (p > u) {
      break;
    }
    long long cnt = 0;
    while (u % p == 0) {
      tot++;
      cnt++;
      u /= p;
    }
    div = div * divmod(factor[cnt]) % MOD;
  }
  if (u > 1) {
    tot++;
  }
  long long res = factor[tot] * div % MOD;
  return mp[key] = res;
}
long long solve(long long D, long long a, long long b) {
  if (a > b) {
    swap(a, b);
  }
  long long g = gcd(a, b);
  long long pa = calcpath(a / g);
  long long pb = calcpath(b / g);
  return pa * pb % MOD;
}
void test() {
  init(12);
  assert(solve(12, 12, 1) == 3);
  assert(solve(12, 4, 4) == 1);
  assert(calcpath(12) == 3);
  mp.clear();
  init(288807105787200LL);
  assert(solve(288807105787200LL, 46LL, 482955026400LL) == 547558588LL);
  mp.clear();
}
template <class T>
inline void scan_d(T &ret) {
  char c;
  ret = 0;
  while ((c = getchar()) < '0' || c > '9')
    ;
  while (c >= '0' && c <= '9') {
    ret = ret * 10 + (c - '0'), c = getchar();
  }
}
int main() {
  long long D;
  scan_d(D);
  init(D);
  int q;
  scan_d(q);
  while (q--) {
    long long a, b;
    scan_d(a);
    scan_d(b);
    puts(to_string(solve(D, a, b)).c_str());
  }
  return 0;
}