#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const int maxn = 1e5 + 10;
long long fac[maxn], ifac[maxn];
long long qpow(long long x, long long n) {
  if (n <= 0) return 1ll;
  long long ret = qpow(x, n >> 1);
  ret = ret * ret % mod;
  if (n & 1) ret = ret * x % mod;
  return ret;
}
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
int main() {
  long long D;
  scanf("%lld", &D);
  int q;
  scanf("%d", &q);
  vector<long long> factors;
  long long M = D;
  for (long long p = 2ll; p * p <= D; ++p) {
    if (M % p == 0) {
      factors.push_back(p);
      while (M % p == 0) {
        M /= p;
      }
    }
  }
  if (M > 1) factors.push_back(M);
  fac[0] = 1;
  ifac[0] = qpow(fac[0], mod - 2);
  for (int i = 1; i < maxn; ++i) {
    fac[i] = fac[i - 1] * i % mod;
    ifac[i] = qpow(fac[i], mod - 2);
  }
  while (q--) {
    long long x, y;
    scanf("%lld %lld", &x, &y);
    long long d = gcd(x, y);
    x /= d;
    y /= d;
    vector<int> a, b;
    for (long long i : factors) {
      long long cnt1 = 0, cnt2 = 0;
      while (x % i == 0) {
        ++cnt1;
        x /= i;
      }
      while (y % i == 0) {
        ++cnt2;
        y /= i;
      }
      if (cnt1) a.push_back(cnt1);
      if (cnt2) b.push_back(cnt2);
    }
    long long ret = 1;
    int s1 = 0, s2 = 0;
    for (int i : a) s1 += i, ret = ret * ifac[i] % mod;
    for (int i : b) s2 += i, ret = ret * ifac[i] % mod;
    ret = ret * fac[s1] % mod;
    ret = ret * fac[s2] % mod;
    printf("%lld\n", ret);
  }
}

### 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;
const long long mod = 998244353;
const int maxn = 1e5 + 10;
long long fac[maxn], ifac[maxn];
long long qpow(long long x, long long n) {
  if (n <= 0) return 1ll;
  long long ret = qpow(x, n >> 1);
  ret = ret * ret % mod;
  if (n & 1) ret = ret * x % mod;
  return ret;
}
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
int main() {
  long long D;
  scanf("%lld", &D);
  int q;
  scanf("%d", &q);
  vector<long long> factors;
  long long M = D;
  for (long long p = 2ll; p * p <= D; ++p) {
    if (M % p == 0) {
      factors.push_back(p);
      while (M % p == 0) {
        M /= p;
      }
    }
  }
  if (M > 1) factors.push_back(M);
  fac[0] = 1;
  ifac[0] = qpow(fac[0], mod - 2);
  for (int i = 1; i < maxn; ++i) {
    fac[i] = fac[i - 1] * i % mod;
    ifac[i] = qpow(fac[i], mod - 2);
  }
  while (q--) {
    long long x, y;
    scanf("%lld %lld", &x, &y);
    long long d = gcd(x, y);
    x /= d;
    y /= d;
    vector<int> a, b;
    for (long long i : factors) {
      long long cnt1 = 0, cnt2 = 0;
      while (x % i == 0) {
        ++cnt1;
        x /= i;
      }
      while (y % i == 0) {
        ++cnt2;
        y /= i;
      }
      if (cnt1) a.push_back(cnt1);
      if (cnt2) b.push_back(cnt2);
    }
    long long ret = 1;
    int s1 = 0, s2 = 0;
    for (int i : a) s1 += i, ret = ret * ifac[i] % mod;
    for (int i : b) s2 += i, ret = ret * ifac[i] % mod;
    ret = ret * fac[s1] % mod;
    ret = ret * fac[s2] % mod;
    printf("%lld\n", ret);
  }
}

```