#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long N = 1e6 + 10;
const long long Bsz = 340;
const long long inf = 1e18;
const long long mod = 998244353;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long d;
long long q;
long long fact[N];
map<long long, long long> dmp;
void factorize(map<long long, long long> &mp, long long first) {
  if (first == 1) return;
  for (long long i = min((long long)sqrt(first) + 1, first - 1); i >= 2; --i)
    if (first % i == 0) {
      factorize(mp, first / i);
      factorize(mp, i);
      return;
    }
  ++mp[first];
}
void factorize2(map<long long, long long> &mp, long long first) {
  for (auto p : dmp)
    while (first % p.first == 0) {
      first /= p.first;
      ++mp[p.first];
    }
}
long long __gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long bpow(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;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  scanf("%lli%lli", &d, &q);
  factorize(dmp, d);
  fact[0] = 1;
  for (long long i = 1; i < 100; ++i) fact[i] = fact[i - 1] * i % mod;
  while (q--) {
    long long a, b;
    scanf("%lli%lli", &a, &b);
    long long gcd = __gcd(a, b);
    a /= gcd;
    b /= gcd;
    if (a < b) swap(a, b);
    map<long long, long long> mpa;
    map<long long, long long> mpb;
    factorize2(mpa, a);
    factorize2(mpb, b);
    long long donthave = 0;
    long long overhave = 0;
    long long have = 0;
    long long del1 = 1;
    long long del2 = 1;
    long long del3 = 1;
    for (const auto &p : mpa) {
      long long cnta = p.second;
      long long cntb = mpb[p.first];
      if (cnta > cntb) {
        overhave += cnta - cntb;
        del1 = del1 * fact[cnta - cntb] % mod;
      } else {
        donthave += cntb - cnta;
        del2 = del2 * fact[cntb - cnta] % mod;
      }
    }
    for (const auto &p : mpb) {
      have += p.second;
      del3 = del3 * fact[p.second] % mod;
    }
    long long ans = fact[donthave] * fact[overhave] % mod * fact[have] % mod *
                    bpow(del1 * del2 % mod * del3 % mod, mod - 2) % mod;
    printf("%lli\n", ans);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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). 

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long N = 1e6 + 10;
const long long Bsz = 340;
const long long inf = 1e18;
const long long mod = 998244353;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long d;
long long q;
long long fact[N];
map<long long, long long> dmp;
void factorize(map<long long, long long> &mp, long long first) {
  if (first == 1) return;
  for (long long i = min((long long)sqrt(first) + 1, first - 1); i >= 2; --i)
    if (first % i == 0) {
      factorize(mp, first / i);
      factorize(mp, i);
      return;
    }
  ++mp[first];
}
void factorize2(map<long long, long long> &mp, long long first) {
  for (auto p : dmp)
    while (first % p.first == 0) {
      first /= p.first;
      ++mp[p.first];
    }
}
long long __gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long bpow(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;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  scanf("%lli%lli", &d, &q);
  factorize(dmp, d);
  fact[0] = 1;
  for (long long i = 1; i < 100; ++i) fact[i] = fact[i - 1] * i % mod;
  while (q--) {
    long long a, b;
    scanf("%lli%lli", &a, &b);
    long long gcd = __gcd(a, b);
    a /= gcd;
    b /= gcd;
    if (a < b) swap(a, b);
    map<long long, long long> mpa;
    map<long long, long long> mpb;
    factorize2(mpa, a);
    factorize2(mpb, b);
    long long donthave = 0;
    long long overhave = 0;
    long long have = 0;
    long long del1 = 1;
    long long del2 = 1;
    long long del3 = 1;
    for (const auto &p : mpa) {
      long long cnta = p.second;
      long long cntb = mpb[p.first];
      if (cnta > cntb) {
        overhave += cnta - cntb;
        del1 = del1 * fact[cnta - cntb] % mod;
      } else {
        donthave += cntb - cnta;
        del2 = del2 * fact[cntb - cnta] % mod;
      }
    }
    for (const auto &p : mpb) {
      have += p.second;
      del3 = del3 * fact[p.second] % mod;
    }
    long long ans = fact[donthave] * fact[overhave] % mod * fact[have] % mod *
                    bpow(del1 * del2 % mod * del3 % mod, mod - 2) % mod;
    printf("%lli\n", ans);
  }
  return 0;
}

```