//26进制与10进制之间的相互转换

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 1e6 + 10;
bool find(string s)
{
	bool x = false, y =false, z = false;
	for(int i = 0; i < s.size(); i ++)
	{
		if(s[0] == 'R')
		{
			x = true;
		}
		if(x == true && s[i] <= '9' && s[i] >= '0')y = true;
		if(y == true && s[i] == 'C')
		{
			return z;
		}
	}
	z = true;
	return z;
}
int main()
{
	int n;
	cin >>n;
	while(n --)
	{
		string s;
		cin >> s;
		if(find(s))
		{
			int l = 0,  r = 0, c = 0;
			for(int i = 0; i < s.size(); i ++)
				if(s[i] >= 'A' && s[i] <= 'Z')l ++;
			int num = 1;
			for(int i = l - 1; i >= 0; i --){c += (s[i] - 'A' + 1) * num, num *= 26;}
			num = 1;
			for(int i = s.size() - 1; i >= l; i --)r += (s[i] -'0') * num, num *= 10;
			printf("R%dC%d\n",r,c); 
		}
		else {
			string ss = "";
			int l = 0, r = 0, c = 0;
			for(int i = 1; l < s.size(); i ++)
				if(s[i] <= '9' && s[i] >= '0')l ++;
				else break;
			int num = 1;
			for(int i = l; i >= 1; i --)r += (s[i] -'0') * num, num *= 10;
			num = 1;
			for(int i = s.size() - 1; i >= l + 2; i --)c += (s[i] - '0') * num, num *= 10;
			while(c > 0)
			{
				if(c % 26 != 0)ss += (c % 26 + 'A' - 1);
				else ss += 'Z', c -= 1;//注意当余数为0时,是Z,而且要 -1, 因为不减的话, 下面一位就会比原来多出来一个
				c /= 26;
			}
			reverse(ss.begin(), ss.end());
			cout << ss<< r <<endl;
		}
	}
}