Problem: This time Baby Ehab will only cut and not stick. He starts with a piece of paper with an array a of length n written on it, and then he does the following:

  * he picks a range (l, r) and cuts the subsegment a_l, a_{l + 1}, …, a_r out, removing the rest of the array. 
  * he then cuts this range into multiple subranges. 
  * to add a number theory spice to it, he requires that the elements of every subrange must have their product equal to their [least common multiple (LCM)](https://en.wikipedia.org/wiki/Least_common_multiple). 



Formally, he partitions the elements of a_l, a_{l + 1}, …, a_r into contiguous subarrays such that the product of every subarray is equal to its LCM. Now, for q independent ranges (l, r), tell Baby Ehab the minimum number of subarrays he needs.

Input

The first line contains 2 integers n and q (1 ≤ n,q ≤ 10^5) — the length of the array a and the number of queries.

The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^5) — the elements of the array a.

Each of the next q lines contains 2 integers l and r (1 ≤ l ≤ r ≤ n) — the endpoints of this query's interval.

Output

For each query, print its answer on a new line.

Example

Input


6 3
2 3 10 7 5 14
1 6
2 4
3 5


Output


3
1
2

Note

The first query asks about the whole array. You can partition it into [2], [3,10,7], and [5,14]. The first subrange has product and LCM equal to 2. The second has product and LCM equal to 210. And the third has product and LCM equal to 70. Another possible partitioning is [2,3], [10,7], and [5,14].

The second query asks about the range (2,4). Its product is equal to its LCM, so you don't need to partition it further.

The last query asks about the range (3,5). You can partition it into [10,7] and [5].

Solution (in PYTHON3):

Solution:

MAXPRIME=100000
primeFactors=[[]for _ in range(MAXPRIME+1)]
for i in range(2,MAXPRIME+1):
    if len(primeFactors[i])==0:
        for j in range(i,MAXPRIME+1,i):
            primeFactors[j].append(i)

def main():
    
    # every element in a subgroup must be co-prime to each other
    n,q=readIntArr()
    arr=readIntArr()
    
    elemPrimes=[inf for _ in range(100001)] # elemPrimes[p] contains latest element with this prime factor
    
    furthestCoprimeIndex=[inf for _ in range(n)] # gives furthest point where all elements in range are coprime to this particular element
    for i in range(n-1,-1,-1):
        for p in primeFactors[arr[i]]:
            furthestCoprimeIndex[i]=min(furthestCoprimeIndex[i],elemPrimes[p]-1)
            elemPrimes[p]=i
    
    furthest2=furthestCoprimeIndex # gives furthest point you can hop to considering all elements in the range
    currmin=inf
    for i in range(n-1,-1,-1):
        furthest2[i]=min(furthest2[i],currmin)
        currmin=min(currmin,furthest2[i])
    
    
    maxPower=18 # 2**17=131072
    dp=[[inf for _ in range(maxPower)] for __ in range(n)] # dp[i][j]=furthest you jump with 2**j jumps
    for i in range(n):
        dp[i][0]=furthest2[i]
    for j in range(1,maxPower):
        for i in range(n):
            if dp[i][j-1]+1>=n:
                break # all subsequent is inf
            else:
                dp[i][j]=dp[dp[i][j-1]+1][j-1]
    
    allans=[]
    for _ in range(q):
        l,r=readIntArr()
        l-=1
        r-=1
        
        r2=l-1
        cnts=0
        while True:
            # keep greedily going as far as possible with all elements being coprime
            l2=r2+1
            breakOuter=False
            for j in range(maxPower):
                if dp[l2][j]>=r:
                    if j==0:
                        cnts+=1
                        breakOuter=True
                    else:
                        cnts+=(1<<(j-1))
                        r2=dp[l2][j-1]
                        # print('l:{} r:{} l2:{} r2:{} cnts:{}'.format(l,r,l2,r2,cnts))
                    break
            if breakOuter:
                break
        allans.append(cnts)
        
    multiLineArrayPrint(allans)
    
    return

import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
 
def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
# def readFloatArr():
#     return [float(x) for x in input().split()]

def makeArr(defaultVal,dimensionArr): # eg. makeArr(0,[n,m])
    dv=defaultVal;da=dimensionArr
    if len(da)==1:return [dv for _ in range(da[0])]
    else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
 
def queryInteractive(x,y):
    print('? {} {}'.format(x,y))
    sys.stdout.flush()
    return int(input())
 
def answerInteractive(ans):
    print('! {}'.format(ans))
    sys.stdout.flush()
 
inf=float('inf')
MOD=10**9+7
# MOD=998244353
 
 
for _abc in range(1):
    main()