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 os
import sys
from io import BytesIO, IOBase

_print = print
BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = 'x' in file.mode or 'r' not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b'\n') + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode('ascii'))
        self.read = lambda: self.buffer.read().decode('ascii')
        self.readline = lambda: self.buffer.readline().decode('ascii')


def inp():
    return sys.stdin.readline().rstrip()


def mpint():
    return map(int, inp().split(' '))


def itg():
    return int(inp())


# ############################## import
_FAC = 10 ** 5


def prime_sieve(n):
    """returns a sieve of primes >= 5 and < n"""
    flag = n % 6 == 2
    sieve = bytearray((n // 3 + flag >> 3) + 1)
    for i in range(1, int(n ** 0.5) // 3 + 1):
        if not (sieve[i >> 3] >> (i & 7)) & 1:
            k = (3 * i + 1) | 1
            for j in range(k * k // 3, n // 3 + flag, 2 * k):
                sieve[j >> 3] |= 1 << (j & 7)
            for j in range(k * (k - 2 * (i & 1) + 4) // 3, n // 3 + flag, 2 * k):
                sieve[j >> 3] |= 1 << (j & 7)
    return sieve


def prime_list(n):
    """returns a list of primes <= n"""
    res = []
    if n > 1:
        res.append(2)
    if n > 2:
        res.append(3)
    if n > 4:
        sieve = prime_sieve(n + 1)
        res.extend(3 * i + 1 | 1 for i in range(1, (n + 1) // 3 + (n % 6 == 1)) if not (sieve[i >> 3] >> (i & 7)) & 1)
    return res


PRIME_FACTORS = [set() for _ in range(_FAC + 1)]
for p in prime_list(_FAC):
    kp = p
    while kp <= _FAC:
        PRIME_FACTORS[kp].add(p)
        kp += p

# ############################## main
EXP = _FAC.bit_length() + 1  # 18, 2**18 > 1e5


def main():
    n, q = mpint()
    arr = tuple(mpint())

    # dp[i][j]: The new index which we need to start at if we performed it 2**i times
    dp = [[0] * n for _ in range(EXP)]

    # dp[0][j]: i.e. the largest j s.t. lcm(arr[i:j]) = reduce(mul, arr[i:j])
    # lcm(arr[i:j]) = reduce(mul, arr[i:j]) iff
    # there is no common prime factor for each pair of number in arr[i:j]

    # Calculating dp[0][j], using two pointers
    i = 0
    st = PRIME_FACTORS[arr[0]].copy()
    for j, a in enumerate(arr[1:], 1):
        pf = PRIME_FACTORS[a]
        cp = st & pf  # common prime
        if cp:  # have common prime factor, need new start
            while cp:
                dp[0][i] = j
                prev_pf = PRIME_FACTORS[arr[i]]
                st -= prev_pf
                cp -= prev_pf
                i += 1
        st |= pf
    # deal with i ~ n-1
    while i != n:
        dp[0][i] = n
        i += 1
    # _print(dp[0])

    # dp doubling
    # dp[i][j] = dp[i-1][dp[i-1][j]]
    # since dp[i-1][j] is the previous starting point of j performed 2**i times
    for i in range(1, EXP):
        for j in range(n):
            lf = dp[i - 1][j]
            dp[i][j] = n if lf == n else dp[i - 1][lf]

    # Now we can performed the starting points quickly
    for _ in range(q):
        lf, rg = mpint()
        lf = dp[0][lf - 1]
        ans = 1  # 2**0 = 1
        while lf < rg:
            for step in range(1, EXP):
                if dp[step][lf] >= rg:
                    # let previous bound be the new start
                    ans += 1 << step - 1
                    lf = dp[step - 1][lf]
                    break
        print(ans)


DEBUG = 0
URL = 'https://codeforces.com/contest/1516/problem/D'

if __name__ == '__main__':
    # 0: normal, 1: runner, 2: debug, 3: interactive
    if DEBUG == 1:
        import requests
        from ACgenerator.Y_Test_Case_Runner import TestCaseRunner

        runner = TestCaseRunner(main, URL)
        inp = runner.input_stream
        print = runner.output_stream
        runner.checking()
    else:
        if DEBUG != 2:
            sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
        if DEBUG:
            def print(*args, **kwargs):
                _print(*args, **kwargs)
                sys.stdout.flush()
        main()
# Please check!