#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 2e6 + 10;
const long long INF = 2e9;
const long long mod = 998244353;
long long D, q, u, v, gcd;
long long a[MAXN], s[MAXN], cnt;
long long fac[MAXN];
long long GCD(long long x, long long y) {
  if (!y) return x;
  return GCD(y, x % y);
}
inline long long ksm(long long x, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) (res *= x) %= mod;
    b >>= 1;
    (x *= x) %= mod;
  }
  return res;
}
signed main() {
  scanf("%lld%lld", &D, &q);
  long long dd = D;
  for (register long long i = 2; i * i <= dd; ++i) {
    if (D % i) continue;
    a[++cnt] = i;
    while (D % i == 0) s[cnt]++, D /= i;
  }
  if (D != 1) a[++cnt] = D, s[cnt] = 1;
  fac[0] = 1;
  for (register long long i = 1; i <= (long long)2e6; ++i)
    fac[i] = (fac[i - 1] * i) % mod;
  while (q--) {
    scanf("%lld%lld", &u, &v);
    if (u == v) {
      puts("1");
      continue;
    }
    gcd = GCD(u, v);
    u /= gcd, v /= gcd;
    long long cntu = 0, cntv = 0, tmpu = 1, tmpv = 1;
    for (register long long i = 1, tmp; i <= cnt; ++i) {
      tmp = 0;
      if (a[i] > u) break;
      while (u % a[i] == 0) tmp++, u /= a[i];
      (tmpu *= fac[tmp]) %= mod;
      cntu += tmp;
    }
    for (register long long i = 1, tmp; i <= cnt; ++i) {
      tmp = 0;
      if (a[i] > v) break;
      while (v % a[i] == 0) tmp++, v /= a[i];
      (tmpv *= fac[tmp]) %= mod;
      cntv += tmp;
    }
    printf("%lld\n", (fac[cntu] * fac[cntv]) % mod * ksm(tmpu, mod - 2) % mod *
                         ksm(tmpv, mod - 2) % mod);
  }
  return 0;
}

### 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>
using namespace std;
const long long MAXN = 2e6 + 10;
const long long INF = 2e9;
const long long mod = 998244353;
long long D, q, u, v, gcd;
long long a[MAXN], s[MAXN], cnt;
long long fac[MAXN];
long long GCD(long long x, long long y) {
  if (!y) return x;
  return GCD(y, x % y);
}
inline long long ksm(long long x, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) (res *= x) %= mod;
    b >>= 1;
    (x *= x) %= mod;
  }
  return res;
}
signed main() {
  scanf("%lld%lld", &D, &q);
  long long dd = D;
  for (register long long i = 2; i * i <= dd; ++i) {
    if (D % i) continue;
    a[++cnt] = i;
    while (D % i == 0) s[cnt]++, D /= i;
  }
  if (D != 1) a[++cnt] = D, s[cnt] = 1;
  fac[0] = 1;
  for (register long long i = 1; i <= (long long)2e6; ++i)
    fac[i] = (fac[i - 1] * i) % mod;
  while (q--) {
    scanf("%lld%lld", &u, &v);
    if (u == v) {
      puts("1");
      continue;
    }
    gcd = GCD(u, v);
    u /= gcd, v /= gcd;
    long long cntu = 0, cntv = 0, tmpu = 1, tmpv = 1;
    for (register long long i = 1, tmp; i <= cnt; ++i) {
      tmp = 0;
      if (a[i] > u) break;
      while (u % a[i] == 0) tmp++, u /= a[i];
      (tmpu *= fac[tmp]) %= mod;
      cntu += tmp;
    }
    for (register long long i = 1, tmp; i <= cnt; ++i) {
      tmp = 0;
      if (a[i] > v) break;
      while (v % a[i] == 0) tmp++, v /= a[i];
      (tmpv *= fac[tmp]) %= mod;
      cntv += tmp;
    }
    printf("%lld\n", (fac[cntu] * fac[cntv]) % mod * ksm(tmpu, mod - 2) % mod *
                         ksm(tmpv, mod - 2) % mod);
  }
  return 0;
}

```