You are given T test cases. For each test case, you have:

• An integer N, the size of an array. • N integers a₁, a₂, …, aₙ.

A “frequency” of a value x is how many times x appears among these N integers. You need to count how many distinct values appear an odd number of times in the array and print that count.

Formally, let distinct(a) be the set of different elements in the array a. If freq(x) denotes the number of times x appears in a, then you must compute the number of x in distinct(a) such that freq(x) is odd.

For each position i (1 ≤ i ≤ N), consider what happens when you remove the element at index i. The resulting array has length N−1, and the former positions j > i “shift” left by one. That is, after removing aᵢ, the array becomes [a₁, a₂, …, aᵢ₋₁, aᵢ₊₁, …, aₙ], and the element originally at position j (where j>i) now occupies position j−1.

We say an index i is “good” if, in this newly formed array, the sum of the elements at the odd positions is equal to the sum of the elements at the even positions (1-based indexing in the new array). You need to count how many such “good” indices exist in the original array.

Formally: • Let b = [b₁, b₂, …, bₙ₋₁] be the array formed after removing aᵢ. • b₁ = a₁, b₂ = a₂, …, bᵢ₋₁ = aᵢ₋₁, and for k ≥ i, bₖ = a₍ᵏ₊₁₎. • i is “good” if ∑(bₖ : k odd) = ∑(bₖ : k even).

You must compute, for each test case, how many “good” indices there are and output that count.