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). 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
template <class A, class B, class C>
struct triple {
  A first;
  B second;
  C third;
  bool operator<(const triple &t) const {
    if (first != t.first) return first < t.first;
    if (second != t.second) return second < t.second;
    return third < t.third;
  }
};
const long double EPS = 0.000000001;
const long double INFD32 = 1000000000;
const long long INF64 = 1000000000000000000;
const long long INF32 = 1000000000;
const long long one = 1;
bool iss_ll(long long __1, long long __2) { return 2 * INF64 / __1 >= __2; }
bool equal(double _1, double _2) { return (abs(_1 - _2) < EPS); }
bool iss_sqr(long long &__1) {
  long long __2 = sqrt((long double)__1);
  return __2 * __2 == __1;
};
bool iss_prime(long long __1) {
  for (long long i = 2; i * i <= __1; ++i)
    if (!(__1 % i)) return 0;
  return 1;
}
long long gcd(long long __1, long long __2) {
  while (__1 && __2)
    if (__1 > __2)
      __1 = __1 % __2;
    else
      __2 = __2 % __1;
  return __1 + __2;
}
long long lcm(long long __1, long long __2) {
  return __1 * __2 / (gcd(__1, __2));
}
long long sq_cube(long long x) {
  if (x == 0) return 1;
  long long l = 0, r = 2e6;
  while (l + 1 < r) {
    long long m = (l + r) / 2;
    if (m * m * m > x)
      r = m;
    else
      l = m;
  }
  return l;
}
long long bp(long long a, long long b, long long md) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    a %= md;
    res %= md;
    b >>= 1;
  }
  return res;
}
long long inv(long long x, long long md) { return bp(x, md - 2, md); }
void update_by_mod(long long &x, long long md) {
  x %= md;
  if (x < 0) x += md;
}
void update_by_mod(long long &x, long long &y, long long md) {
  update_by_mod(x, md);
  update_by_mod(y, md);
}
void update_by_mod(long long &x, long long &y, long long &z, long long md) {
  update_by_mod(x, md);
  update_by_mod(y, md);
  update_by_mod(z, md);
}
void update_by_mod(long long &x, long long &y, long long &z, long long &a,
                   long long md) {
  update_by_mod(x, md);
  update_by_mod(y, md);
  update_by_mod(z, md);
  update_by_mod(a, md);
}
vector<long long> z_function(string &s) {
  long long n = ((long long)((s).size()));
  vector<long long> ans(n, 0);
  long long l = 0, r = 0;
  for (long long i = 1; i < n; ++i) {
    ans[i] = min(r - i + 1, ans[i - l]);
    while (ans[i] + i < n && s[ans[i]] == s[i + ans[i]]) {
      ++ans[i];
    }
    if (i + ans[i] - 1 > r) {
      l = i, r = i + ans[i] - 1;
    }
    r = max(r, i);
  }
  return ans;
}
long long a[500500], b[500500];
long long md = 998244353;
long long fact[500500];
vector<long long> dels;
void solve() {
  long long d;
  cin >> d;
  for (long long i = 2; i * i <= d; ++i) {
    if (d % i == 0) {
      dels.push_back(i);
      while (d % i == 0) d /= i;
    }
  }
  if (d > 1) dels.push_back(d);
  long long q;
  cin >> q;
  fact[0] = 1;
  for (long long i = 1; i < 400; ++i) fact[i] = fact[i - 1] * i % md;
  while (q--) {
    long long a, b;
    cin >> a >> b;
    long long cnt1 = 0, cnt2 = 0;
    vector<long long> v1, v2;
    for (long long i : dels) {
      if (a % i == 0 || b % i == 0) {
        long long cn1 = 0, cn2 = 0;
        while (a % i == 0) {
          ++cn1;
          a /= i;
        }
        while (b % i == 0) {
          ++cn2;
          b /= i;
        }
        if (cn1 == cn2) continue;
        if (cn1 > cn2)
          v1.push_back(cn1 - cn2), cnt1 += cn1 - cn2;
        else
          v2.push_back(cn2 - cn1), cnt2 += cn2 - cn1;
      }
    }
    long long res1 = fact[cnt1], res2 = fact[cnt2];
    for (auto i : v1) {
      res1 *= inv(fact[i], md);
      res1 %= md;
    }
    for (auto i : v2) {
      res2 *= inv(fact[i], md);
      res2 %= md;
    }
    cout << res1 * res2 % md << "\n";
  }
}
signed main(void) {
  srand(time(0));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}