def main():
    import sys
    input = sys.stdin.read
    data = input().split()
 
    index = 0
    t = int(data[index])
    index += 1
    results = []
 
    for _ in range(t):
        n = int(data[index])
        index += 1
        a = list(map(int, data[index:index + n]))
        index += n
 
        count = 0
        diff_count = {}  # 用于统计差值出现的次数
 
        for i in range(n):
            # 计算当前元素与位置（i+1）的差值
            # 原题中i,j是数学上的1-based索引，所以这里用i+1
            curr_diff = a[i] - (i + 1)
 
            # 之前已经有多个相同差值的元素，每个都能与当前元素形成有效对
            if curr_diff in diff_count:
                count += diff_count[curr_diff]
 
            # 更新当前差值的出现次数（放在后面处理，确保i<j的顺序）
            diff_count[curr_diff] = diff_count.get(curr_diff, 0) + 1
 
        results.append(str(count))
 
    sys.stdout.write("\n".join(results) + "\n")
 
if __name__ == "__main__":
    main()

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4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6