1334_E. Divisor Paths
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). 
```python
import sys
input = sys.stdin.buffer.readline
mod=998244353
d=int(input())
div=[]
x=2
di=0
t=0
while d>1 and x*x<=d:
    if d%x==0:
        div.append(x)
        d//=x
        di+=1
        t+=1
    while d%x==0:
        d//=x
        t+=1
    x+=1
if d>1:
    div.append(d)
    di+=1
    t+=1
fac=[1]
for i in range(1,t+1):
    fac.append(i*fac[-1]%mod)
for f in range(int(input())):
    v,u=map(int,input().split())
    divs1=[]
    divs2=[]
    tot1=0
    tot2=0
    for i in range(di):
        while v%div[i]==0 and u%div[i]==0:
            v//=div[i]
            u//=div[i]
        if v%div[i]==0:
            divs1.append(0)
        if u%div[i]==0:
            divs2.append(0)
        while v%div[i]==0:
            v//=div[i]
            divs1[-1]+=1
            tot1+=1
        while u%div[i]==0:
            u//=div[i]
            divs2[-1]+=1
            tot2+=1
    res=fac[tot1]*fac[tot2]
    tod=1
    for x in divs1:
        tod*=fac[x]
        tod%=mod
    for x in divs2:
        tod*=fac[x]
        tod%=mod
    res*=pow(tod,mod-2,mod)
    res%=mod
    print(res)
```