#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using pii = pair<long long, long long>;
const bool debug = 0;
const long long MAX_N = 1020;
const long long mod = 998244353;
long long modpow(long long x, long long y, long long p = mod) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long fact[MAX_N + 5];
long long invfact[MAX_N + 5];
long long binom(long long n, long long r) {
  long long t = (fact[n] * (invfact[r])) % mod;
  return (t * invfact[n - r]) % mod;
}
void precompute() {
  fact[0] = 1;
  invfact[0] = 1;
  for (long long i = 1; i <= MAX_N; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= mod;
  }
  invfact[MAX_N] = modpow(fact[MAX_N], mod - 2);
  for (long long i = MAX_N - 1; i >= 1; i--) {
    invfact[i] = invfact[i + 1] * (i + 1);
    invfact[i] %= mod;
  }
}
long long frac(long long a, long long b) {
  return (a * (modpow(b, mod - 2))) % mod;
}
long long D;
vector<long long> pdivlist;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  precompute();
  cin >> D;
  while (D % 2 == 0) {
    pdivlist.push_back(2);
    D /= 2;
  }
  for (long long i = 3; i <= sqrt(D); i += 2) {
    while (D % i == 0) {
      pdivlist.push_back(i);
      D /= i;
    }
  }
  if (D >= 2) pdivlist.push_back(D);
  pdivlist.erase(unique(((pdivlist).begin()), ((pdivlist).end())),
                 pdivlist.end());
  long long q;
  cin >> q;
  long long pp = pdivlist.size();
  while (q--) {
    vector<long long> a(pp, 0), b(pp, 0);
    long long x, y;
    cin >> x >> y;
    for (long long i = 0; i < pp; i++) {
      while (x % pdivlist[i] == 0) {
        x /= pdivlist[i];
        a[i]++;
      }
      while (y % pdivlist[i] == 0) {
        y /= pdivlist[i];
        b[i]++;
      }
    }
    vector<long long> neg, pos;
    long long ns = 0, ps = 0;
    long long ans = 1;
    for (long long i = 0; i < pp; i++) {
      long long k = (a[i] - b[i]);
      if (k < 0) {
        k = -k;
        ns += k;
        neg.push_back(k);
      } else if (k > 0) {
        ps += k;
        pos.push_back(k);
      }
    }
    ans = fact[ns] * fact[ps];
    ans %= mod;
    for (auto it : neg) {
      ans *= invfact[it];
      ans %= mod;
    }
    for (auto it : pos) {
      ans *= invfact[it];
      ans %= mod;
    }
    ans %= mod;
    cout << ans << '\n';
  }
}

### Prompt

Please formulate a cpp solution to the following 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). 

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using pii = pair<long long, long long>;
const bool debug = 0;
const long long MAX_N = 1020;
const long long mod = 998244353;
long long modpow(long long x, long long y, long long p = mod) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long fact[MAX_N + 5];
long long invfact[MAX_N + 5];
long long binom(long long n, long long r) {
  long long t = (fact[n] * (invfact[r])) % mod;
  return (t * invfact[n - r]) % mod;
}
void precompute() {
  fact[0] = 1;
  invfact[0] = 1;
  for (long long i = 1; i <= MAX_N; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= mod;
  }
  invfact[MAX_N] = modpow(fact[MAX_N], mod - 2);
  for (long long i = MAX_N - 1; i >= 1; i--) {
    invfact[i] = invfact[i + 1] * (i + 1);
    invfact[i] %= mod;
  }
}
long long frac(long long a, long long b) {
  return (a * (modpow(b, mod - 2))) % mod;
}
long long D;
vector<long long> pdivlist;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  precompute();
  cin >> D;
  while (D % 2 == 0) {
    pdivlist.push_back(2);
    D /= 2;
  }
  for (long long i = 3; i <= sqrt(D); i += 2) {
    while (D % i == 0) {
      pdivlist.push_back(i);
      D /= i;
    }
  }
  if (D >= 2) pdivlist.push_back(D);
  pdivlist.erase(unique(((pdivlist).begin()), ((pdivlist).end())),
                 pdivlist.end());
  long long q;
  cin >> q;
  long long pp = pdivlist.size();
  while (q--) {
    vector<long long> a(pp, 0), b(pp, 0);
    long long x, y;
    cin >> x >> y;
    for (long long i = 0; i < pp; i++) {
      while (x % pdivlist[i] == 0) {
        x /= pdivlist[i];
        a[i]++;
      }
      while (y % pdivlist[i] == 0) {
        y /= pdivlist[i];
        b[i]++;
      }
    }
    vector<long long> neg, pos;
    long long ns = 0, ps = 0;
    long long ans = 1;
    for (long long i = 0; i < pp; i++) {
      long long k = (a[i] - b[i]);
      if (k < 0) {
        k = -k;
        ns += k;
        neg.push_back(k);
      } else if (k > 0) {
        ps += k;
        pos.push_back(k);
      }
    }
    ans = fact[ns] * fact[ps];
    ans %= mod;
    for (auto it : neg) {
      ans *= invfact[it];
      ans %= mod;
    }
    for (auto it : pos) {
      ans *= invfact[it];
      ans %= mod;
    }
    ans %= mod;
    cout << ans << '\n';
  }
}

```