human
Consider a sequence of digits of length 2^k [a_1, a_2, …, a_{2^k}]. We perform the following operation with it: replace pairs (a_{2i+1}, a_{2i+2}) with (a_{2i+1} + a_{2i+2})mod 10 for 0≤ i<2^{k-1}. For every i where a_{2i+1} + a_{2i+2}≥ 10 we get a candy! As a result, we will get a sequence of length 2^{k-1}.

Less formally, we partition sequence of length 2^k into 2^{k-1} pairs, each consisting of 2 numbers: the first pair consists of the first and second numbers, the second of the third and fourth …, the last pair consists of the (2^k-1)-th and (2^k)-th numbers. For every pair such that sum of numbers in it is at least 10, we get a candy. After that, we replace every pair of numbers with a remainder of the division of their sum by 10 (and don't change the order of the numbers).

Perform this operation with a resulting array until it becomes of length 1. Let f([a_1, a_2, …, a_{2^k}]) denote the number of candies we get in this process. 

For example: if the starting sequence is [8, 7, 3, 1, 7, 0, 9, 4] then:

After the first operation the sequence becomes [(8 + 7)mod 10, (3 + 1)mod 10, (7 + 0)mod 10, (9 + 4)mod 10] = [5, 4, 7, 3], and we get 2 candies as 8 + 7 ≥ 10 and 9 + 4 ≥ 10.

After the second operation the sequence becomes [(5 + 4)mod 10, (7 + 3)mod 10] = [9, 0], and we get one more candy as 7 + 3 ≥ 10. 

After the final operation sequence becomes [(9 + 0) mod 10] = [9]. 

Therefore, f([8, 7, 3, 1, 7, 0, 9, 4]) = 3 as we got 3 candies in total.

You are given a sequence of digits of length n s_1, s_2, … s_n. You have to answer q queries of the form (l_i, r_i), where for i-th query you have to output f([s_{l_i}, s_{l_i+1}, …, s_{r_i}]). It is guaranteed that r_i-l_i+1 is of form 2^k for some nonnegative integer k.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the sequence.

The second line contains n digits s_1, s_2, …, s_n (0 ≤ s_i ≤ 9).

The third line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.

Each of the next q lines contains two integers...
The above is tricky. Write me a correct solution in PYTHON3.

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What is a the Python code you requested?
assistant
#Here is the Python code you requested.

import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline

class SegmentTree():
    def __init__(self, n, func, init=float('inf')):
        self.n = 2**(n-1).bit_length()
        self.init = init
        self.data = [init]*(2*self.n)
        self.func = func

    def set(self, k, v):
        self.data[k+self.n-1] = v

    def build(self):
        for k in reversed(range(self.n-1)):
            self.data[k] = self.func(self.data[k*2+1], self.data[k*2+2])

    def query(self, l, r):
        L = l+self.n
        R = r+self.n
        ret = self.init
        while L < R:
            if R & 1:
                R -= 1
                ret = self.func(ret, self.data[R-1])
            if L & 1:
                ret = self.func(ret, self.data[L-1])
                L += 1
            L >>= 1
            R >>= 1
        return ret

N = int(input())
L = list(map(int, input().split()))
Q = int(input())
Query = [list(map(lambda x:int(x)-1, input().split())) for _ in range(Q)]

def func(a, b):
    return a+b

Seg = SegmentTree(N, func, init=0)
for k, v in enumerate(L):
    Seg.set(k, v)
Seg.build()

for query in Query:
    l, r = query
    a = Seg.query(l, r+1)
    print(a//10)