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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class d {
	public static void main(String[] args) {
		FastScanner scan=new FastScanner();
		PrintWriter out=new PrintWriter(System.out);
		
		/*
		consider the following case:
		6 1
		5 2 2 2 5 5
		1 6
		
		the answer is 4 even though the max number of items w/ shared factors in the array is 3.
		therefore my previous methodology was incorrect :(
		
		for the subarray to be good, all the elements in it must be coprime.
		it's always best to greedily take the largest subarray you can and go like that til you reach the end
		since if you don't, it will only potentially hurt you in the future.
		
		now let's consider for each index i, a value skip[i] that is the lowest index that breaks your current array
		i.e. we skip directly there to start a new array. if we can figure out these values, we can do binary lifting
		to answer the queries quickly!
		
		we should be able to do a two pointer sweep for this
		and maintain frequencies of each factor
		*/
		
		int n=scan.nextInt(), q=scan.nextInt();
		
		boolean[] prime=new boolean[100001];
		Arrays.fill(prime,true);
		prime[0]=prime[1]=false;
		
		for(int i=2;i<prime.length;i++) {
			if(!prime[i]) continue;
			for(int j=i+i;j<prime.length;j+=i) {
				prime[j]=false;
			}
		}
		
		ArrayList<Integer>[] facs=new ArrayList[n];
		for(int i=0;i<n;i++) facs[i]=new ArrayList<>();
		
		int[] a=new int[n];
		for(int i=0;i<n;i++) {
			a[i]=scan.nextInt();
			int cur=a[i];
			for(int j=2;j*j<=100000;j++) {
				if(cur%j==0&&prime[j]) {
					facs[i].add(j);
					while(cur%j==0) cur/=j;
				}
				int other=cur/j;
				if(cur%j==0&&cur%other==0&&prime[other]) {
					facs[i].add(other);
					while(cur%other==0) cur/=other;
				}
			}
			if(cur!=1) facs[i].add(cur);
		}
		
		int[][] skip=new int[20][n];
		for(int[] s:skip) Arrays.fill(s,n);
		
		int l=0, r=0;
		int[] f=new int[100001];
		int ctExceeded=0;
		
		while(r<n) {
			while(r<n&&ctExceeded==0) {
				for(int factor:facs[r]) {
					f[factor]++;
					if(f[factor]>1) ctExceeded++;
				}
				r++;
			}
			if(ctExceeded>0) {
				while(ctExceeded>0) {
					skip[0][l]=r-1;
					for(int factor:facs[l]) {
						if(f[factor]>1) ctExceeded--;
						f[factor]--;
					}
					l++;
				}
			}
		}
		
		for(int j=1;j<20;j++) {
			for(int i=0;i<n;i++) {
				if(skip[j-1][i]==n) skip[j][i]=n;
				else skip[j][i]=skip[j-1][skip[j-1][i]];
			}
		}
		
		for(int i=0;i<q;i++) {
			int ql=scan.nextInt()-1, qr=scan.nextInt()-1;
			int res=0, cur=ql;
			for(int j=19;j>=0;j--) {
				if(skip[j][cur]<=qr) {
					cur=skip[j][cur];
					res+=(1<<j);
				}
			}
			//one more step to get *beyond* qr
			out.println(res+1);
		}
		
		out.close();
	}
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			try	{
				br = new BufferedReader(new InputStreamReader(System.in));
				st = new StringTokenizer(br.readLine());
			} catch (Exception e){e.printStackTrace();}
		}

		public String next() {
			if (st.hasMoreTokens())	return st.nextToken();
			try {st = new StringTokenizer(br.readLine());}
			catch (Exception e) {e.printStackTrace();}
			return st.nextToken();
		}

		public int nextInt() {return Integer.parseInt(next());}

		public long nextLong() {return Long.parseLong(next());}

		public double nextDouble() {return Double.parseDouble(next());}

		public String nextLine() {
			String line = "";
			if(st.hasMoreTokens()) line = st.nextToken();
			else try {return br.readLine();}catch(IOException e){e.printStackTrace();}
			while(st.hasMoreTokens()) line += " "+st.nextToken();
			return line;
		}
	}
}