B. Spreadsheets Codeforces Beta Round 1

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该编程问题要求将给定的Excel单元格坐标，如A1到AZ100000和R1C1到R100000C100000，按照另一种编号系统（例如RXCY）进行转换。

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 n lines, 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.

Examples

input

Copy

2
R23C55
BC23

output

Copy

BC23
R23C55

题目大致意思：给两种日期记录模式，让我们相互转换。

日期记录模式一：日期一、字母（26进制，不超过26不进位），日期二、数字。

日期记录模式二：日期二、"R"+数字，日期一、"C"+数字。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"

void solve(){
	string s;
	cin >> s;
	ll x,y,a=0,b=0;
	bool f=true,pd=false;
	map<char,ll>mp;
	for(ll i = 0 ; i < s.size() ;  i++)
		if(s[i] >= 'A' && s[i] <= 'Z')mp[s[i]]++;
	if(mp['C'] != 1 || mp['R'] != 1 || mp.size() != 2)f=false;
	if(f)
		for(ll i = 1 ; i < s.size() ; i++)
			if(s[i]>='A' && s[i] <='Z')
				if(s[i-1] >= '0' && s[i-1] <= '9')pd=true;
	if(f && pd){
		x=s.find("R");
		y=s.rfind("C");
		for(ll i = 1 ; i < s.size() ; i ++){
			if(i < y)a=a*10+s[i]-'0';
			else if(i > y)b=b*10+s[i]-'0';
		}
		s="";
		while(b){
			if(b%26 == 0){
				s='Z'+s;
				b--;
			}else s=char('A'+b%26-1)+s;
			b/=26;
		}
		cout << s << a << endl;
	}else{
		for(ll i = 0 ; i < s.size() ; i ++){
			if(s[i] >= 'A' && s[i] <= 'Z')a=a*26+s[i]-'A'+1;
			else b=b*10+s[i]-'0';
		}
		cout << "R" << b << "C" << a << endl;
	}
	return;
}

int main(){
	ll t=1;cin >> t;
	while(t--)solve();
	return 0;
}