## 题目描述

Little X has met the following problem recently.

Let’s define $f(x)$ as the sum of digits in decimal representation of number $x$ (for example, $f(1234)=1+2+3+4$). You are to calculate $\sum_{i=l}^r f(i) \bmod a$.

Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:

```
ans = solve(l, r) % a;
if (ans <= 0) ans += a;
```

This code will fail only on the test with $\sum_{i=l}^r f(i) \equiv 0 \pmod a$. You are given number $a$, help Little X to find a proper test for hack.

## 题意概述

令$f(i)$表示$i$各位数字之和。给定$a$，要求找出一对$(l, r)$满足$\sum_{i=l}^r f(i) \equiv 0 \pmod a$。

数据范围：$1 \le a \le 10^{18}$。

## 算法分析

思路很巧妙。有一个显而易见的结论：$\forall i \in [1, 10^{18}], \; f(i)+1=f(i+10^{18})$。它有一个推论：$\forall x \in [1, 10^{18}], \; 1+\sum_{i=x}^{x+10^{18}-1} f(i)=\sum_{i=x+1}^{x+10^{18}} f(i)$。因此只要算出$t=\sum_{i=1}^{10^{18}} f(i) \bmod a$，将区间右移$a-t$单位。

## 代码

#include <cstdio> long long mul(long long a, long long b, long long mod) { long long ret = 0; while (b) b & 1 && ((ret += a) %= mod), (a <<= 1) %= mod, b >>= 1; return ret; } int main() { long long n; scanf("%lld", &n); long long cur = mul(1800000000000000000ll, 45, n) + 1, l = 1, r = 1000000000000000000ll; l += n - cur, r += n - cur, printf("%lld %lld\n", l, r); return 0; }