You are given an array of N elements and num queries, In each query you are given three numbers L,R and K and you have to tell, how many indexes are there in between L and R(L<=i<=R) such that the frequency of a[i] from index i to n-1 is k.
Note: 0-based indexing
Example 1:
Input:
N=5
num=3
A={1,1,3,4,3}
Q={{0,2,2},{0,2,1},{0,4,2}}
Output:
2 1 2
Explanation:
For query 1: 0 2 2
L=0,R=2,K=2
let, L<=i<=R
For i=0: frequency of a[i] i.e. 1 from i to n-1 is 2.
For i=1: frequency of a[i] i.e. 1 from i to n-1 is 1.
For i=2: frequency of a[i] i.e. 3 from i to n-1 is 2.
Hence we have two elements from index 0 to 2
whose frequency from i to n-1 is 2.
For query 2: 0 2 1
L=0,R=2,K=1
As we can see from the above query that there is
only a single element in 0 to 2 whose frequency
from i to n-1 is 1.
For query 3: 0 4 2
The answer will be 2 because of the index 0 and 2.
Example 2:
Input:
N=5
num=2
A={1,1,1,1,1}
Q={{0,4,2},{0,4,1}}
Output:
1 1
Explanation:
For query 1: 0 4 2
L=0,R=4,K=2
let, L<=i<=R
For i=0: frequency of a[i] i.e. 1 from i to n-1 is 5.
For i=1: frequency of a[i] i.e. 1 from i to n-1 is 4.
For i=2: frequency of a[i] i.e. 1 from i to n-1 is 3.
For i=3: frequency of a[i] i.e. 1 from i to n-1 is 2.
For i=4: frequency of a[i] i.e. 1 from i to n-1 is 1.
Hence we have one elements from index 0 to 4 whose frequency from i to n-1 is 2.
Similarly For query 2:
there is only 1 element in 0 to 4 whose frequency from i to n-1 is 1.
Expected Time Complexity: O(N^{2})
Expected Auxiliary Space: O(N^{2})
Your Task:
You don't need to read input or print anything. Your task is to complete the function solveQueries() which take two variable N and num representing the length of the original array and number of queries and two arrays as input, A and Q representing the original array and an array of queries(2-D array with 3 columns of L,R and K respectively), and returns the array of length num with the solution to each query.

Constraints:
1 <= N <= 10^{3}
0 <= Q < 10^{3}
1 <= A[i] <= 10^{5}

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