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...
Solve the task in PYTHON3.

Give me a the Python program you requested?
assistant
#Here is the Python program you requested.

def naiveSolve():
    
    
    
    return

def main():
    
    n=int(input())
    s=readIntArr()
    
    maxN=1
    while 2**maxN<n:
        maxN+=1
    rem=makeArr(lambda:0,[n,maxN])
    val=makeArr(lambda:0,[n,maxN+1])
    
    for i in range(n):
        rem[i][0]=s[i]
        val[i][0]=0
    for i in range(n-1,-1,-1):
        for j in range(1,maxN):
            if i+(2**j)-1>=n:
                break
            rem[i][j]=(rem[i][j-1]+rem[i+2**(j-1)][j-1])%10
    for i in range(n-1,-1,-1):
        for j in range(1,maxN+1):
            if i+(2**j)-1>=n:
                break
            val[i][j]=val[i][j-1]+val[i+2**(j-1)][j-1]
            if rem[i][j-1]+rem[i+2**(j-1)][j-1]>=10:
                val[i][j]+=1
    
    allans=[]
    q=int(input())
    for _ in range(q):
        l,r=readIntArr()
        j=0
        while 2**j<(r-l+1):
            j+=1
        allans.append(val[l-1][j])
    multiLineArrayPrint(allans)
    
    return

import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.

def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
# def readFloatArr():
#     return [float(x) for x in input().split()]
 
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
    dv=defaultValFactory;da=dimensionArr
    if len(da)==1:return [dv() for _ in range(da[0])]
    else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
 
def queryInteractive(l,r):
    print('? {} {}'.format(l,r))
    sys.stdout.flush()
    return int(input())
 
def answerInteractive(x):
    print('! {}'.format(x))
    sys.stdout.flush()
 
inf=float('inf')
# MOD=10**9+7
MOD=998244353

from math import gcd,floor,ceil
# from math import...