Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...

Your task is to print the k-th digit of this sequence.

Input

The first and only line contains integer k (1 ≤ k ≤ 10000) — the position to process (1-based index).

Output

Print the k-th digit of the resulting infinite sequence.

Examples

Input


7


Output


7


Input


21


Output


5
# Aaditya Upadhyay
# ..............
# ╭━┳━╭━╭━╮╮
# ┃┈┈┈┣▅╋▅┫┃
# ┃┈┃┈╰━╰━━━━━━╮
# ╰┳╯┈┈┈┈┈┈┈┈┈◢▉◣
# ╲┃┈┈┈┈┈┈┈┈┈┈▉▉▉
# ╲┃┈┈┈┈┈┈┈┈┈┈◥▉◤
# ╲┃┈┈┈┈╭━┳━━━━╯
# ╲┣━━━━━━┫
# ……….
# .……. /´¯/)………….(\¯`\
# …………/….//……….. …\….\
# ………/….//……………....\….\
# …./´¯/…./´¯\……/¯ `\…..\¯`\
# ././…/…/…./|_…|.\….\….\…\.\
# (.(….(….(…./.)..)...(.\.).).)
# .\…………….\/../…....\….\/…………/
# ..\…………….. /……...\………………../
# …..\…………… (………....)……………./
from sys import stdin, stdout
from collections import *
from math import gcd, floor, ceil
def st(): return list(stdin.readline().strip())


def li(): return list(map(int, stdin.readline().split()))
def mp(): return map(int, stdin.readline().split())
def inp(): return int(stdin.readline())
def pr(n): return stdout.write(str(n)+"\n")


mod = 1000000007
INF = float('inf')


def solve():
    n = inp() - 1
    pre = 0
    cur = 1
    point = 1
    while pre + 9*cur*point < n:
        pre += 9*cur*point
        cur *= 10
        point += 1
    n -= pre
    num, pos = divmod(n, point)
    num += pow(10, point-1)
    pr(str(num)[pos])


for _ in range(1):
    solve()
