#include <bits/stdc++.h>
using namespace std;
struct Primality {
  vector<long long> A = {2, 3, 5, 7, 11, 13, 17, 19, 23};
  long long largemul(long long a, long long b, long long n) {
    long long r = 0;
    for (; b; b >>= 1, a <<= 1) {
      if (a >= n) a -= n;
      if (b & 1) {
        r += a;
        if (r >= n) r -= n;
      }
    }
    return r;
  }
  long long fastexp(long long a, long long b, long long n) {
    long long ret = 1;
    for (; b; b >>= 1, a = largemul(a, a, n))
      if (b & 1) ret = largemul(ret, a, n);
    return ret;
  }
  bool mrtest(long long n) {
    if (n == 1) return false;
    long long d = n - 1;
    int s = 0;
    while ((d & 1) == 0) {
      s++;
      d >>= 1;
    }
    s--;
    if (s < 0) s = 0;
    for (int j = 0; j < (int)A.size(); j++) {
      if (A[j] >= n) continue;
      long long ad = fastexp(A[j], d, n);
      if (ad == 1) continue;
      bool notcomp = false;
      long long a2rd = ad;
      for (int r = 0; r <= s; r++) {
        if (a2rd == n - 1) {
          notcomp = true;
          break;
        }
        a2rd = largemul(a2rd, a2rd, n);
      }
      if (!notcomp) {
        return false;
      }
    }
    return true;
  }
  long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
  long long pollard_rho(long long n) {
    int i = 0, k = 2;
    long long x = 3, y = 3;
    while (1) {
      i++;
      x = largemul(x, x, n) - 1;
      if (x < 0) x += n;
      long long d = gcd(llabs(y - x), n);
      if (d != 1 && d != n) return d;
      if (i == k) y = x, k <<= 1;
    }
  }
};
const int N = 1e6 + 6, S = 505, mod = 998244353;
long long C[S][S];
int main() {
  for (int i = 0; i < S; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; ++j) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
  }
  Primality np;
  long long d;
  scanf("%lld", &d);
  vector<long long> primes;
  for (int x = 2; x < N; ++x) {
    if (d % x) continue;
    while ((d % x) == 0) {
      d /= x;
    }
    primes.push_back(x);
  }
  if (d > 1) {
    if (np.mrtest(d)) {
      primes.push_back(d);
    } else {
      long long x = np.pollard_rho(d);
      long long y = d / y;
      primes.push_back(x);
      if (y != x) primes.push_back(y);
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    long long u, v;
    scanf("%lld %lld", &u, &v);
    long long g = np.gcd(u, v);
    u /= g;
    v /= g;
    int lu = 0, lv = 0;
    long long ans = 1;
    for (long long p : primes) {
      int cu = 0, cv = 0;
      while ((u % p) == 0) {
        ++cu;
        u /= p;
      }
      while ((v % p) == 0) {
        ++cv;
        v /= p;
      }
      if (cu > 0) ans = ans * C[lu + cu][cu] % mod;
      if (cv > 0) ans = ans * C[lv + cv][cv] % mod;
      lu += cu;
      lv += cv;
    }
    ans %= mod;
    if (ans < 0) ans += mod;
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Generate 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>
using namespace std;
struct Primality {
  vector<long long> A = {2, 3, 5, 7, 11, 13, 17, 19, 23};
  long long largemul(long long a, long long b, long long n) {
    long long r = 0;
    for (; b; b >>= 1, a <<= 1) {
      if (a >= n) a -= n;
      if (b & 1) {
        r += a;
        if (r >= n) r -= n;
      }
    }
    return r;
  }
  long long fastexp(long long a, long long b, long long n) {
    long long ret = 1;
    for (; b; b >>= 1, a = largemul(a, a, n))
      if (b & 1) ret = largemul(ret, a, n);
    return ret;
  }
  bool mrtest(long long n) {
    if (n == 1) return false;
    long long d = n - 1;
    int s = 0;
    while ((d & 1) == 0) {
      s++;
      d >>= 1;
    }
    s--;
    if (s < 0) s = 0;
    for (int j = 0; j < (int)A.size(); j++) {
      if (A[j] >= n) continue;
      long long ad = fastexp(A[j], d, n);
      if (ad == 1) continue;
      bool notcomp = false;
      long long a2rd = ad;
      for (int r = 0; r <= s; r++) {
        if (a2rd == n - 1) {
          notcomp = true;
          break;
        }
        a2rd = largemul(a2rd, a2rd, n);
      }
      if (!notcomp) {
        return false;
      }
    }
    return true;
  }
  long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
  long long pollard_rho(long long n) {
    int i = 0, k = 2;
    long long x = 3, y = 3;
    while (1) {
      i++;
      x = largemul(x, x, n) - 1;
      if (x < 0) x += n;
      long long d = gcd(llabs(y - x), n);
      if (d != 1 && d != n) return d;
      if (i == k) y = x, k <<= 1;
    }
  }
};
const int N = 1e6 + 6, S = 505, mod = 998244353;
long long C[S][S];
int main() {
  for (int i = 0; i < S; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; ++j) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
  }
  Primality np;
  long long d;
  scanf("%lld", &d);
  vector<long long> primes;
  for (int x = 2; x < N; ++x) {
    if (d % x) continue;
    while ((d % x) == 0) {
      d /= x;
    }
    primes.push_back(x);
  }
  if (d > 1) {
    if (np.mrtest(d)) {
      primes.push_back(d);
    } else {
      long long x = np.pollard_rho(d);
      long long y = d / y;
      primes.push_back(x);
      if (y != x) primes.push_back(y);
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    long long u, v;
    scanf("%lld %lld", &u, &v);
    long long g = np.gcd(u, v);
    u /= g;
    v /= g;
    int lu = 0, lv = 0;
    long long ans = 1;
    for (long long p : primes) {
      int cu = 0, cv = 0;
      while ((u % p) == 0) {
        ++cu;
        u /= p;
      }
      while ((v % p) == 0) {
        ++cv;
        v /= p;
      }
      if (cu > 0) ans = ans * C[lu + cu][cu] % mod;
      if (cv > 0) ans = ans * C[lv + cv][cv] % mod;
      lu += cu;
      lv += cv;
    }
    ans %= mod;
    if (ans < 0) ans += mod;
    printf("%lld\n", ans);
  }
  return 0;
}

```