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:
'''
import sys
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
import bisect,string,math,time,functools,random,fractions
from bisect import*
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
rep=range;R=range
def I():return int(input())
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def AI():return map(int,open(0).read().split())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
def RI(a=1,b=10):return random.randint(a,b)
def INP():
    N=10
    n=random.randint(1,N)
    a=RLI(n,0,n-1)
    return n,a
def Rtest(T):
    case,err=0,0
    for i in range(T):
        inp=INP()
        a1=naive(*inp)
        a2=solve(*inp)
        if a1!=a2:
            print(inp)
            print('naive',a1)
            print('solve',a2)
            err+=1
        case+=1
    print('Tested',case,'case with',err,'errors')
def GI(V,E,ls=None,Directed=False,index=1):
    org_inp=[];g=[[] for i in range(V)]
    FromStdin=True if ls==None else False
    for i in range(E):
        if FromStdin:
            inp=LI()
            org_inp.append(inp)
        else:
            inp=ls[i]
        if len(inp)==2:a,b=inp;c=1
        else:a,b,c=inp
        if index==1:a-=1;b-=1
        aa=(a,c);bb=(b,c);g[a].append(bb)
        if not Directed:g[b].append(aa)
    return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
    #h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
    mp=[boundary]*(w+2);found={}
    for i in R(h):
        s=input()
        for char in search:
            if char in s:
                found[char]=((i+1)*(w+2)+s.index(char)+1)
                mp_def[char]=mp_def[replacement_of_found]
        mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
    mp+=[boundary]*(w+2)
    return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def accum(ls):
    rt=[0]
    for i in ls:rt+=[rt[-1]+i]
    return rt
def bit_combination(n,base=2):
    rt=[]
    for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
    return rt
def gcd(x,y):
    if y==0:return x
    if x%y==0:return y
    while x%y!=0:x,y=y,x%y
    return y
def YN(x):print(['NO','YES'][x])
def Yn(x):print(['No','Yes'][x])
def show(*inp,end='\n'):
    if show_flg:print(*inp,end=end)

mo=10**9+7
#mo=998244353
inf=float('inf')
FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
alp=[chr(ord('a')+i)for i in range(26)]
#sys.setrecursionlimit(10**7)
def gcj(c,x):
    print("Case #{0}:".format(c+1),x)
'''
########################################################################################################################################################################
# Verified by
# https://yukicoder.me/problems/no/979
# https://atcoder.jp/contests/abc152/tasks/abc152_e

## return prime factors of N as dictionary {prime p:power of p}
## within 2 sec for N = 2*10**20+7
def primeFactor(N):
    i,n=2,N
    ret={}
    d,sq=2,99
    while i<=sq:
        k=0
        while n%i==0:
            n,k,ret[i]=n//i,k+1,k+1
        if k>0 or i==97:
            sq=int(n**(1/2)+0.5)
        if i<4:
            i=i*2-1
        else:
            i,d=i+d,d^6
    if n>1:
        ret[n]=1
    return ret

## return divisors of n as list
def divisor(n):
    div=[1]
    for i,j in primeFactor(n).items():
        div=[(i**k)*d for d in div for k in range(j+1)]
    return div

## return the list of prime numbers in [2,N], using eratosthenes sieve
## around 800 ms for N = 10**6  by PyPy3 (7.3.0) @ AtCoder
def PrimeNumSet(N):
    M=int(N**0.5)
    seachList=[i for i in range(2,N+1)]
    primes=[]
    while seachList:
        if seachList[0]>M:
            break
        primes.append(seachList[0])
        tmp=seachList[0]
        seachList=[i for i in seachList if i%tmp!=0]
    return primes+seachList


## retrun LCM of numbers in list b
## within 2sec for no of B = 10*5  and  Bi < 10**6
def LCM(b,mo=10**9+7):
    prs=PrimeNumSet(max(b))
    M=dict(zip(prs,[0]*len(prs)))
    for i in b:
        dc=primeFactor(i)
        for j,k in dc.items():
            M[j]=max(M[j],k)
    
    r=1
    for j,k in M.items():
        if k!=0:
            r*=pow(j,k,mo)
            r%=mo
    return r

## return (a,b,gcd(x,y)) s.t. a*x+b*y=gcd(x,y)
def extgcd(x,y):
    if y==0:
        return 1,0
    r0,r1,s0,s1 = x,y,1,0
    while r1!= 0:
        r0,r1,s0,s1=r1,r0%r1,s1,s0-r0//r1*s1
    return s0,(r0-s0*x)//y,x*s0+y*(r0-s0*x)//y


## return x,LCM(mods) s.t. x = rem_i (mod_i), x = -1 if such x doesn't exist
## verified by ABC193E
## https://atcoder.jp/contests/abc193/tasks/abc193_e
def crt(rems,mods):
    n=len(rems)
    if n!=len(mods):
        return NotImplemented
    x,d=0,1
    
    for r,m in zip(rems,mods):
        a,b,g=extgcd(d,m)
        x,d=(m*b*x+d*a*r)//g,d*(m//g)
        x%=d

    for r,m in zip(rems,mods):
        if r!=x%m:
            return -1,d

    return x,d

## returns the maximum integer rt s.t. rt*rt<=x
## verified by ABC191D
## https://atcoder.jp/contests/abc191/tasks/abc191_d
def intsqrt(x):
    if x<0:
        return NotImplemented
    rt=int(x**0.5)-1
    while (rt+1)**2<=x:
        rt+=1
    return rt


show_flg=False
show_flg=True

ans=0


import sys
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
from collections import deque,defaultdict,Counter
def I():return int(input())
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]

n,q=LI()
a=LI()
pf=dict()

d=dict()
for i in set(a):
    pf[i]=primeFactor(i).items()

l,r=0,0
div=defaultdict(int)
a+=a[-1],
nx=[n]*(n+1)

while r<=n:
    if all([div[p]==0 for p,cnt in pf[a[r]]]): # tmp*s[r]<=k:  # 条件を満たすならrを1進める
        for p,cnt in pf[a[r]]:
            div[p]+=cnt
        r+=1
    else:
        nx[l]=r # 次の仕切りはrの前
        if l<r:  # 条件を満たさず、l<rならlを1進める
            for p,cnt in pf[a[l]]:
                div[p]-=cnt
            l+=1
        else:  # lがrを追い越さないように。l=rならrを進める。
            for p,cnt in pf[a[r]]:
                div[p]+=cnt
            r+=1    
    #if r<n:
    #    show((l,r),a[r],div)
    

nx=[nx]

# doubling
m=n.bit_length()
for k in range(m):
    nnx=[0]*(n+1)
    for i in range(n+1):
        nnx[i]=nx[k][nx[k][i]]
    nx.append(nnx)
    #show(nx,nnx)


for i in range(q):
    l,r=LI_()
    
    t=1
    for p in range(m,-1,-1):
        if nx[p][l]<=r:
            t+=1<<p
            l=nx[p][l]
    
    print(t)