#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = double;
using pii = pair<ll, ll>;
using vi = vector<ll>;
const ll P = 998244353;
ll n_, n, q, u, v, fac[1111111], inv[1111111];
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % P;
    a = a * a % P;
    b >>= 1;
  }
  return ret;
}
ll c(ll n, ll m) { return fac[n] * inv[m] % P * inv[n - m] % P; }
signed main() {
  scanf("%lld %lld", &n_, &q);
  n = n_;
  vi pr;
  for (ll i = 2; i * i <= n; i++)
    if (n % i == 0) {
      pr.push_back(i);
      while (n % i == 0) n /= i;
    }
  fac[0] = 1;
  for (ll i = 1; i <= 1000000; i++) fac[i] = fac[i - 1] * i % P;
  inv[1000000] = qpow(fac[1000000], P - 2);
  for (ll i = 1000000; i; i--) inv[i - 1] = inv[i] * i % P;
  if (n != 1) pr.push_back(n);
  while (q--) {
    scanf("%lld %lld", &u, &v);
    vi eu, ev;
    for (ll i : pr) {
      eu.push_back(0), ev.push_back(0);
      while (u % i == 0) u /= i, eu.back()++;
      while (v % i == 0) v /= i, ev.back()++;
    }
    ll sum = 0, ans = 1;
    for (ll i = 0; i < eu.size(); i++) sum += max(0ll, ev[i] - eu[i]);
    for (ll i = 0; i < eu.size(); i++)
      if (ev[i] > eu[i])
        ans = ans * c(sum, ev[i] - eu[i]) % P, sum -= ev[i] - eu[i];
    sum = 0;
    for (ll i = 0; i < eu.size(); i++) sum += max(0ll, eu[i] - ev[i]);
    for (ll i = 0; i < eu.size(); i++)
      if (eu[i] > ev[i])
        ans = ans * c(sum, eu[i] - ev[i]) % P, sum -= eu[i] - ev[i];
    printf("%lld\n", ans);
  }
  fprintf(stderr, "time=%.4f\n", (db)clock() / CLOCKS_PER_SEC);
  return 0;
}

### Prompt

Your challenge is to write 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;
using ll = long long;
using db = double;
using pii = pair<ll, ll>;
using vi = vector<ll>;
const ll P = 998244353;
ll n_, n, q, u, v, fac[1111111], inv[1111111];
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % P;
    a = a * a % P;
    b >>= 1;
  }
  return ret;
}
ll c(ll n, ll m) { return fac[n] * inv[m] % P * inv[n - m] % P; }
signed main() {
  scanf("%lld %lld", &n_, &q);
  n = n_;
  vi pr;
  for (ll i = 2; i * i <= n; i++)
    if (n % i == 0) {
      pr.push_back(i);
      while (n % i == 0) n /= i;
    }
  fac[0] = 1;
  for (ll i = 1; i <= 1000000; i++) fac[i] = fac[i - 1] * i % P;
  inv[1000000] = qpow(fac[1000000], P - 2);
  for (ll i = 1000000; i; i--) inv[i - 1] = inv[i] * i % P;
  if (n != 1) pr.push_back(n);
  while (q--) {
    scanf("%lld %lld", &u, &v);
    vi eu, ev;
    for (ll i : pr) {
      eu.push_back(0), ev.push_back(0);
      while (u % i == 0) u /= i, eu.back()++;
      while (v % i == 0) v /= i, ev.back()++;
    }
    ll sum = 0, ans = 1;
    for (ll i = 0; i < eu.size(); i++) sum += max(0ll, ev[i] - eu[i]);
    for (ll i = 0; i < eu.size(); i++)
      if (ev[i] > eu[i])
        ans = ans * c(sum, ev[i] - eu[i]) % P, sum -= ev[i] - eu[i];
    sum = 0;
    for (ll i = 0; i < eu.size(); i++) sum += max(0ll, eu[i] - ev[i]);
    for (ll i = 0; i < eu.size(); i++)
      if (eu[i] > ev[i])
        ans = ans * c(sum, eu[i] - ev[i]) % P, sum -= eu[i] - ev[i];
    printf("%lld\n", ans);
  }
  fprintf(stderr, "time=%.4f\n", (db)clock() / CLOCKS_PER_SEC);
  return 0;
}

```