#include <bits/stdc++.h>
using namespace std;
const long double e = 0.00000005;
const long double pi = 3.1415926535898;
inline int getint() {
  int val = 0;
  char c;
  while ((c = getchar()) && !(c >= '0' && c <= '9'))
    ;
  do {
    val = (val * 10) + c - '0';
  } while ((c = getchar()) && (c >= '0' && c <= '9'));
  return val;
}
long long gcd(long long i, long long j) {
  if (j == 0)
    return i;
  else
    return gcd(j, i % j);
}
const int Y = 100100;
const long long INF = 10000000000000000;
const long long m = 998244353;
long long fact[Y];
long long inv[Y];
bool dg[40000000];
long long pw(int b, long long et) {
  if (b == 1) return et;
  long long tmp = pw(b / 2, et);
  tmp = (tmp * tmp) % m;
  if (b % 2 == 1) tmp = (tmp * et) % m;
  return tmp;
}
vector<long long> simple;
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  long long D;
  cin >> D;
  int q;
  cin >> q;
  fact[0] = 1;
  long long d = sqrtl(D);
  while (d * d > D) --d;
  while ((d + 1) * (d + 1) <= D) ++d;
  for (long long i = 1; i < Y; ++i) {
    fact[i] = fact[i - 1] * i % m;
  }
  for (long long i = 2; i <= d; ++i) {
    if (D % i == 0) {
      simple.push_back(i);
      while (D % i == 0) D /= i;
    }
  }
  if (D != 1) simple.push_back(D);
  while (q--) {
    long long u, v;
    cin >> u >> v;
    long long g = gcd(max(u, v), min(u, v));
    u /= g;
    v /= g;
    int sm1 = 0, sm2 = 0;
    long long mn = 1;
    for (auto x : simple) {
      if (v % x == 0) {
        int cc = 0;
        while (v % x == 0) {
          ++cc;
          v /= x;
        }
        mn = (mn * fact[cc]) % m;
        sm1 += cc;
      }
      if (u % x == 0) {
        int cc = 0;
        while (u % x == 0) {
          ++cc;
          u /= x;
        }
        mn = (mn * fact[cc]) % m;
        sm2 += cc;
      }
    }
    long long rrr = (fact[sm1] * fact[sm2]) % m;
    long long rr = (((fact[sm1] * fact[sm2]) % m) * pw(m - 2, mn)) % m;
    cout << rr << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
using namespace std;
const long double e = 0.00000005;
const long double pi = 3.1415926535898;
inline int getint() {
  int val = 0;
  char c;
  while ((c = getchar()) && !(c >= '0' && c <= '9'))
    ;
  do {
    val = (val * 10) + c - '0';
  } while ((c = getchar()) && (c >= '0' && c <= '9'));
  return val;
}
long long gcd(long long i, long long j) {
  if (j == 0)
    return i;
  else
    return gcd(j, i % j);
}
const int Y = 100100;
const long long INF = 10000000000000000;
const long long m = 998244353;
long long fact[Y];
long long inv[Y];
bool dg[40000000];
long long pw(int b, long long et) {
  if (b == 1) return et;
  long long tmp = pw(b / 2, et);
  tmp = (tmp * tmp) % m;
  if (b % 2 == 1) tmp = (tmp * et) % m;
  return tmp;
}
vector<long long> simple;
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  long long D;
  cin >> D;
  int q;
  cin >> q;
  fact[0] = 1;
  long long d = sqrtl(D);
  while (d * d > D) --d;
  while ((d + 1) * (d + 1) <= D) ++d;
  for (long long i = 1; i < Y; ++i) {
    fact[i] = fact[i - 1] * i % m;
  }
  for (long long i = 2; i <= d; ++i) {
    if (D % i == 0) {
      simple.push_back(i);
      while (D % i == 0) D /= i;
    }
  }
  if (D != 1) simple.push_back(D);
  while (q--) {
    long long u, v;
    cin >> u >> v;
    long long g = gcd(max(u, v), min(u, v));
    u /= g;
    v /= g;
    int sm1 = 0, sm2 = 0;
    long long mn = 1;
    for (auto x : simple) {
      if (v % x == 0) {
        int cc = 0;
        while (v % x == 0) {
          ++cc;
          v /= x;
        }
        mn = (mn * fact[cc]) % m;
        sm1 += cc;
      }
      if (u % x == 0) {
        int cc = 0;
        while (u % x == 0) {
          ++cc;
          u /= x;
        }
        mn = (mn * fact[cc]) % m;
        sm2 += cc;
      }
    }
    long long rrr = (fact[sm1] * fact[sm2]) % m;
    long long rr = (((fact[sm1] * fact[sm2]) % m) * pw(m - 2, mn)) % m;
    cout << rr << "\n";
  }
  return 0;
}

```