• Concatenate A after B to obtain a new string C. For example, if A = "^__^" and B = "T.T", then C = BA = "T.T^__^".
• Let A = B, B = C -- as the example above A = "T.T", B = "T.T^__^".

Input

The input contains multiple test cases, each contains only one integer N (1 <= N <= 2^63 - 1). Proceed to the end of file.

Output

For each test case, print one character on each line, which is the N-th (index begins with 1) character of this infinite string.

Sample Input

1
2
4
8

Sample Output

T
.
^
T

题意：开始时有两个字符串，A = "^__^" (four characters),  B = "T.T" (three characters)，然后重复执行C=BA，A=B,B = C，问第n个字符是什么。

分析：因为最终的字符串是从前往后递推出来的，所以再求解时，我们可以把这个过程逆过去，直到字符串的长度不超过7，输出即可。

#include<iostream>
#include<string>
using namespace std;
typedef long long LL;
LL a[95];  //保存字符串的长度
int main()
{
    string C = "T.T^__^";
    a[0] = 4;
    a[1] = 3;
    int i;
    for(i = 2; i < 90; i++)
        a[i] = a[i-1] + a[i-2];
    LL n;
    while(cin >> n)
    {
        while(n > 7)
        {
            int pos = lower_bound(a, a+89, n) - a;
            n -= a[pos-1];
        }
        cout << C[n-1] << endl;
    }
    return 0;
}