#include <bits/stdc++.h>
using namespace std;
const long long MAX_N = 2e5 + 7;
long long mod = 998244353;
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long calc(long long a, long long b, vector<long long> &div) {
  vector<pair<long long, long long> > d_a, d_b;
  long long sum = 0;
  vector<long long> x;
  for (long long d : div) {
    d_a.push_back({d, 0});
    d_b.push_back({d, 0});
    while (a % d == 0) {
      d_a[d_a.size() - 1].second++;
      a /= d;
    }
    while (b % d == 0) {
      d_b[d_b.size() - 1].second++;
      b /= d;
    }
    long long c = d_a[d_a.size() - 1].second - d_b[d_b.size() - 1].second;
    sum += d_a[d_a.size() - 1].second - d_b[d_b.size() - 1].second;
    if (c != 0) x.push_back(c);
  }
  sort(x.begin(), x.end());
  long long zn = 1;
  long long fact = 1;
  long long pos = 0;
  for (long long i = 1; i <= sum; ++i) {
    fact *= i;
    fact %= mod;
    while (pos < x.size() && x[pos] == i) {
      zn *= fact;
      zn %= mod;
      pos++;
    }
  }
  long long x1, y1;
  gcd(zn, mod, x1, y1);
  x1 %= mod;
  x1 += mod;
  x1 %= mod;
  return (x1 * fact) % mod;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, q;
  cin >> n >> q;
  vector<long long> div;
  long long r = n;
  for (long long i = 2; i <= sqrt(n); ++i) {
    if (r % i == 0) {
      div.push_back(i);
      while (r % i == 0) r /= i;
    }
  }
  if (r != 1) div.push_back(r);
  while (q--) {
    long long a, b;
    cin >> a >> b;
    long long x1, y1;
    long long d = gcd(a, b, x1, y1);
    cout << (calc(a, d, div) * calc(b, d, div)) % mod << endl;
  }
}

### 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 long MAX_N = 2e5 + 7;
long long mod = 998244353;
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long calc(long long a, long long b, vector<long long> &div) {
  vector<pair<long long, long long> > d_a, d_b;
  long long sum = 0;
  vector<long long> x;
  for (long long d : div) {
    d_a.push_back({d, 0});
    d_b.push_back({d, 0});
    while (a % d == 0) {
      d_a[d_a.size() - 1].second++;
      a /= d;
    }
    while (b % d == 0) {
      d_b[d_b.size() - 1].second++;
      b /= d;
    }
    long long c = d_a[d_a.size() - 1].second - d_b[d_b.size() - 1].second;
    sum += d_a[d_a.size() - 1].second - d_b[d_b.size() - 1].second;
    if (c != 0) x.push_back(c);
  }
  sort(x.begin(), x.end());
  long long zn = 1;
  long long fact = 1;
  long long pos = 0;
  for (long long i = 1; i <= sum; ++i) {
    fact *= i;
    fact %= mod;
    while (pos < x.size() && x[pos] == i) {
      zn *= fact;
      zn %= mod;
      pos++;
    }
  }
  long long x1, y1;
  gcd(zn, mod, x1, y1);
  x1 %= mod;
  x1 += mod;
  x1 %= mod;
  return (x1 * fact) % mod;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, q;
  cin >> n >> q;
  vector<long long> div;
  long long r = n;
  for (long long i = 2; i <= sqrt(n); ++i) {
    if (r % i == 0) {
      div.push_back(i);
      while (r % i == 0) r /= i;
    }
  }
  if (r != 1) div.push_back(r);
  while (q--) {
    long long a, b;
    cin >> a >> b;
    long long x1, y1;
    long long d = gcd(a, b, x1, y1);
    cout << (calc(a, d, div) * calc(b, d, div)) % mod << endl;
  }
}

```