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本文介绍了一个简单的程序，用于实现Excel中两种不同的单元格坐标表示方式之间的转换：一种是以字母标记列号的方式，另一种是以RXCY格式标记行列的方式。程序通过解析输入的坐标，将它们从一种格式转换为另一种格式。

In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.

The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.

Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.

Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.

Input

The first line of the input contains integer number n (1 ≤ n ≤ 10⁵), the number of coordinates in the test. Then there follow nlines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 10⁶ .

Output

Write n lines, each line should contain a cell coordinates in the other numeration system.

Sample test(s)

input
2
R23C55
BC23

output
BC23
R23C55

解题说明：此题就是实现两种表示类型的转换，主要是字符串处理问题

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;

char line[1024], t[1024], *p;
int r, c;

void printcol(int c) 
{
        if(!c) 
		{
			return;
		}
        printcol((c - 1)/26);
        putchar(((c - 1)% 26) + 'A');
}

int main()
{
	int n,i;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%s", line);
        if(sscanf(line,"R%dC%d", &r, &c) == 2) 
		{
            printcol(c);
            printf("%d\n", r);
        } 
		else 
		{
           sscanf(line, "%[A-Z]%d", t, &r);
           for(p = t, c = 0; *p; p++) 
			{
				c = c * 26, c += *p - 'A' + 1;
			}
			printf("R%dC%d\n", r, c);
         }
	}
	return 0;
}