Candies!
Consider a sequence of digits of length $$$2^k$$$ $$$[a_1, a_2, \ldots, a_{2^k}]$$$. We perform the following operation with it: replace pairs $$$(a_{2i+1}, a_{2i+2})$$$ with $$$(a_{2i+1} + a_{2i+2})\bmod 10$$$ for $$$0\le i<2^{k-1}$$$. For every $$$i$$$ where $$$a_{2i+1} + a_{2i+2}\ge 10$$$ 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 $$$\ldots$$$, 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 $$$f([a_1, a_2, \ldots, a_{2^k}])$$$ 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 $$$[(8 + 7)\bmod 10, (3 + 1)\bmod 10, (7 + 0)\bmod 10, (9 + 4)\bmod 10]$$$ $$$=$$$ $$$[5, 4, 7, 3]$$$, and we get $$$2$$$ candies as $$$8 + 7 \ge 10$$$ and $$$9 + 4 \ge 10$$$.After the second operation the sequence becomes $$$[(5 + 4)\bmod 10, (7 + 3)\bmod 10]$$$ $$$=$$$ $$$[9, 0]$$$, and we get one more candy as $$$7 + 3 \ge 10$$$. After the final operation sequence becomes $$$[(9 + 0) \bmod 10]$$$ $$$=$$$ $$$[9]$$$. 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 $$$n$$$ $$$s_1, s_2, \ldots s_n$$$. You have to answer $$$q$$$ queries of the form $$$(l_i, r_i)$$$, where for $$$i$$$-th query you have to output $$$f([s_{l_i}, s_{l_i+1}, \ldots, s_{r_i}])$$$. It is guaranteed that $$$r_i-l_i+1$$$ is of form $$$2^k$$$ for some nonnegative integer $$$k$$$.
The first example illustrates an example from the statement.$$$f([7, 3, 1, 7]) = 1$$$: sequence of operations is $$$[7, 3, 1, 7] \to [(7 + 3)\bmod 10, (1 + 7)\bmod 10]$$$ $$$=$$$ $$$[0, 8]$$$ and one candy as $$$7 + 3 \ge 10$$$ $$$\to$$$ $$$[(0 + 8) \bmod 10]$$$ $$$=$$$ $$$[8]$$$, so we get only $$$1$$$ candy.$$$f([9]) = 0$$$ as we don't perform operations with it.
The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the sequence.The second line contains $$$n$$$ digits $$$s_1, s_2, \ldots, s_n$$$ ($$$0 \le s_i \le 9$$$).The third line contains a single integer $$$q$$$ ($$$1 \le q \le 10^5$$$) — the number of queries.Each of the next $$$q$$$ lines contains two integers $$$l_i$$$, $$$r_i$$$ ($$$1 \le l_i \le r_i \le n$$$) — $$$i$$$-th query. It is guaranteed that $$$r_i-l_i+1$$$ is a nonnegative integer power of $$$2$$$.
Output $$$q$$$ lines, in $$$i$$$-th line output single integer — $$$f([s_{l_i}, s_{l_i + 1}, \ldots, s_{r_i}])$$$, answer to the $$$i$$$-th query.
import sys
p2D = lambda x: print(*x, sep='\n')

def II():
    return int(sys.stdin.buffer.readline())

def MI():
    return map(int, sys.stdin.buffer.readline().split())

def LI():
    return list(map(int, sys.stdin.buffer.readline().split()))

def LLI(rows_number):
    return [LI() for _ in range(rows_number)]

def BI():
    return sys.stdin.buffer.readline().rstrip()

def SI():
    return sys.stdin.buffer.readline().rstrip().decode()

def li():
    return [int(i) for i in input().split()]

def lli(rows):
    return [li() for _ in range(rows)]

def si():
    return input()

def ii():
    return int(input())

def ins():
    return input().split()

def solve():
    n = II()
    s = LI()
    ac = [0]
    for i in range(n):
        ac.append(ac[-1] + s[i])
    for _ in range(II()):
        (l, r) = LI()
        sys.stdout.write(str((ac[r] - ac[l - 1]) // 10) + '\n')

def main():
    solve()
if __name__ == '__main__':
    main()
n = int(input())
from sys import stdin
a = lambda : stdin.readline().split()
lst = [*map(int, a())]
result = ''
lst.insert(0, 0)
for (i, x) in enumerate(lst[1:]):
    lst[i + 1] += lst[i]
q = int(input())
for _ in range(q):
    (l, r) = map(int, a())
    result += str((lst[r] - lst[l - 1]) // 10) + '\n'
print(result[:-1])
n = int(input())
s = list(map(int, input().split()))
soma = [0] * (n + 1)
for i in range(n):
    soma[i + 1] = soma[i] + s[i]
q = int(input())
left = []
right = []
for i in range(q):
    (l, r) = map(int, input().split())
    left.append(l)
    right.append(r)
for i in range(q):
    num = (soma[right[i]] - soma[left[i] - 1]) // 10
    print(num)