Codeforces 1B

B. Spreadsheets

time limit per test

10 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

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

2
R23C55
BC23

output

BC23
R23C55

题意：给出n个字符串，字符串满足以下两种形式中的随机一种：第一种是一串英文加一串数字，如样例中的BC23，表示表格的行号是23，列号是BC，当然，我们规定A,B,C,...,Z分别表示1,2,3,...,26，AA,AB,AC,...,AZ,BA,...,ZZY,ZZZ,AAAA等由此类推分别代表27,28,29,...,52,53,...我们需要把它改为RxCy的形式，其中x就等于原给定的数字，y是通过字母推算得到的数字；同理，第二种就是第一种的逆操作。求出这n个字符串改变形式后的序列。

题解：大力模拟一发。。。因为刚开始出现R可能有两种情况，我们判断第二位，如果第二位为字母，那么一定属于题意中描述的第一种操作；否则如果第二位是数字，还可能有两种情况，如果接下来都是数字，属于第一种操作，否则属于第二种操作，大力模拟一发即可。

#include<bits/stdc++.h>
using namespace std;
char s[100];
int n,f[6],ans[10];
int main()
{
    scanf("%d",&n);
    f[0]=1;
    for (int i=1;i<=5;++i)
    f[i]=f[i-1]*26;
    for (int i=1;i<=n;++i)
    {
        bool flag=0,vis=0;
        int a=0,b=0,pos=0;
        scanf("%s",s);
        int len=strlen(s);
        if (s[0]=='R'&&s[1]>='0'&&s[1]<='9')
        {
            vis=1;
            for (int j=1;j<len;++j)
            if (s[j]>'9')
            {
                pos=j;
                flag=1;
                break;
            }
            if (!flag)
            {
                for (int j=1;j<len;++j)
                b=b*10+s[j]-'0';
                printf("R%dC%d
",b,'R'-'A'+1);
            }
            else
            {
                for (int j=1;j<pos;++j)
                a=a*10+s[j]-'0';
                for (int j=pos+1;j<len;++j)
                b=b*10+s[j]-'0';
                int j;
                for (j=1;b>=f[j];b-=f[j],++j);
                if(b==0)
                {
                    j--;
                    while (j--) printf("Z");
                }
                else
                {
                    --b;
                    for (int k=1;k<=j;++k)
                    {
                        ans[k]=b%26;
                        b/=26;
                    }
                    for (int k=j;k;--k)
                    printf("%c",ans[k]+'A');
                }
                printf("%d
",a);
            }
        }
        if (!vis)
        {
            for (int j=0;j<len;++j)
            if (s[j]>='0'&&s[j]<='9')
            {
                if (!pos) pos=j;
                break;
            }
            printf("R");
            for (int j=pos;j<len;++j)
            printf("%c",s[j]);
            int sum=0;
            for (int j=1;j<pos;++j)
            sum+=f[j];
            for (int j=0;j<pos;++j)
            sum+=f[pos-j-1]*(s[j]-'A');
            ++sum;
            printf("C%d
",sum);
        }
    }
    return 0;
}