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>
using namespace std;
struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
struct ListNode {
  int val;
  ListNode* next;
  ListNode(int x) : val(x), next(NULL) {}
};
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long nxtLong() {
  long long x;
  cin >> x;
  return x;
}
string nxtStr() {
  string x;
  cin >> x;
  return x;
}
char nxtChar() {
  char x;
  cin >> x;
  return x;
}
double nxtDouble() {
  double x;
  cin >> x;
  return x;
}
long double nxtLDouble() {
  long double x;
  cin >> x;
  return x;
}
TreeNode* makeTree(int n) {
  queue<TreeNode*> tree_nodes;
  TreeNode* root;
  if (n) {
    root = new TreeNode(nxt());
    tree_nodes.push(root);
  } else {
    return NULL;
  }
  while (!tree_nodes.empty()) {
    TreeNode* node = tree_nodes.front();
    tree_nodes.pop();
    bool left;
    cin >> left;
    if (left) {
      node->left = new TreeNode(nxt());
      tree_nodes.push(node->left);
    }
    bool right;
    cin >> right;
    if (right) {
      node->right = new TreeNode(nxt());
      tree_nodes.push(node->right);
    }
  }
  return root;
}
ListNode* makeLinkedList(int n) {
  ListNode* list = new ListNode(0);
  ListNode* dummy = list;
  for (int i = 0; i < n; ++i) {
    list->next = new ListNode(nxt());
    list = list->next;
  }
  return dummy->next;
}
const int MOD = 998244353;
int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  if (a < 0) a += MOD;
  return a;
}
int mul(int a, int b) { return a * 1ll * b % MOD; }
int binpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
void solve() {
  long long d;
  scanf("%lld", &d);
  int q;
  scanf("%d", &q);
  vector<long long> primes;
  for (long long i = 2; i * i <= d; ++i)
    if (d % i == 0) {
      primes.push_back(i);
      while (d % i == 0) d /= i;
    }
  if (d > 1) {
    primes.push_back(d);
  }
  vector<int> fact(100), rfact(100);
  fact[0] = 1;
  for (int i = 1; i < 100; ++i) fact[i] = mul(fact[i - 1], i);
  rfact[99] = binpow(fact[99], MOD - 2);
  for (int i = 98; i >= 0; --i) rfact[i] = mul(rfact[i + 1], i + 1);
  for (int i = 0; i < int(q); i++) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    vector<int> up, dw;
    for (auto p : primes) {
      int cnt = 0;
      while (x % p == 0) {
        --cnt;
        x /= p;
      }
      while (y % p == 0) {
        ++cnt;
        y /= p;
      }
      if (cnt < 0)
        dw.push_back(-cnt);
      else if (cnt > 0)
        up.push_back(cnt);
    }
    int ans = 1;
    ans = mul(ans, fact[accumulate(up.begin(), up.end(), 0)]);
    for (auto it : up) ans = mul(ans, rfact[it]);
    ans = mul(ans, fact[accumulate(dw.begin(), dw.end(), 0)]);
    for (auto it : dw) ans = mul(ans, rfact[it]);
    printf("%d\n", ans);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(9);
  solve();
  return 0;
}