output_description: For each getSum operation, print the sum in a line.
input_description: n q
      com1 x1 y1
      com2 x2 y2
      ...
      comq xq yq
      
  
  In the first line, n (the number of elements in A) and q (the number of
  queries) are given. Then, q queries are given where com represents the type of
  queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
attempt: class RSQ(object):
      INIT = 0
  
      def __init__(self, num) -> None:
          n = 1
          while n <= num:
              n = n * 2
          self.n = n
          self.val = [self.INIT] * (2 * n - 1)
  
      @staticmethod
      def merge(a: int, b: int) -> int:
          return a + b
  
      def update(self, k: int, x: int) -> None:
          k = k + self.n - 1
          self.val[k] += x
          while k:
              k = (k - 1) // 2
              self.val[k] = self.merge(self.val[k * 2 + 1], self.val[k * 2 + 2])
  
      def dfs(self, beg: int, end: int, k: int, l: int, r: int) -> int:
          if end <= l or r <= beg:
              return self.INIT
          if beg <= l and r <= end:
              return self.val[k]
          vl = self.dfs(beg, end, k * 2 + 1, l, (l + r) // 2)
          vr = self.dfs(beg, end, k * 2 + 2, (l + r) // 2, r)
          return self.merge(vl, vr)
  
      def query(self, beg: int, end: int) -> int:
          return self.dfs(beg, end, 0, 0, self.n)
  
  
  if __name__ == "__main__":
      n, q = map(lambda x: int(x), input().split())
      rmq = RSQ(n)
  
      for _ in range(q):
          com, x, y = map(lambda x: int(x), input().split())
          if 0 == com:
              rmq.update(x, y)
          else:
              print(rmq.query(x, y + 1))
  
problem_description: Range Sum Query
  
  Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the
  following operations:
  
    * add(i, x): add x to ai.
    * getSum(s, t): print the sum of as, as+1,...,at.
  
  Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
samples: [{"input": "3 5\n    0 1 1\n    0 2 2\n    0 3 3\n    1 1 2\n    1 2 2", "output": "3\n    2"}]