B. Spreadsheets

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
inputCopy
2
R23C55
BC23
outputCopy
BC23
R23C55

The solution to the above question is:-

#include <bits/stdc++.h>

using namespace std;

#define int long long int
#define F first
#define S second
#define pb push_back
#define si set<int>
#define vi vector<int>
#define pii pair<int, int>
#define vpi vector<pii>
#define vpp vector<pair<int, pii>>
#define mii map<int, int>
#define mpi map<pii, int>
#define spi set<pii>
#define endl "\n"
#define sz(x) ((int)x.size())
#define all(p) p.begin(), p.end()
#define double long double
#define que_max priority_queue<int>
#define que_min priority_queue<int, vi, greater<int>>
#define bug(...) __f(#__VA_ARGS__, __VA_ARGS__)
#define print(a)          \
    for (auto x : a)      \
        cout << x << " "; \
    cout << endl
#define print1(a)    \
    for (auto x : a) \
    cout << x.F << " " << x.S << endl
#define print2(a, x, y)         \
    for (int i = x; i < y; i++) \
        cout << a[i] << " ";    \
    cout << endl

inline int power(int a, int b)
{
    int x = 1;
    while (b)
    {
        if (b & 1)
            x *= a;
        a *= a;
        b >>= 1;
    }
    return x;
}

template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args)
{
    const char *comma = strchr(names + 1, ',');
    cout.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
}

const int N = 200005;

string convertToNumber(string s){
    return "";
}

string convertToString(string s){
    string ans="";
    int num=stoi(s);
    while(num>0){
        char c='Z';
        int curr=num%26;
        char cc=curr+'A'-1;
        if(curr==0){
            ans+='Z';
            num--;
        }else{
                        ans+=cc;
        }
        num/=26;
    }
    reverse(ans.begin(),ans.end());
    return ans;
}


void solve()
{

    string s;
    cin >> s;
    bool first = false;
    int n=s.length();

    for(int i=2; i<n; i++){
        if(s[i]=='C')first=true;
    }
    if(s[0]=='R' && first && s[1]<='9'){
        string ans="";
        int pos=-1;
        for(int i=0; i<n; i++){
            if(s[i]=='C')pos=i;
        }
        string row=s.substr(1,pos-1);
        string col=s.substr(pos+1,n-1);
        string ss=convertToString(col);
        cout<<ss+row<<endl;
        
    }
    else{
        string col="";
        string row="";
        for(int i=0; i<n; i++){
            if(s[i]>='A' && s[i]<='Z'){
                col+=s[i];
            }
            else{
                row+=s[i];
            }
        }
        int column=0;
        int pow=1;
        for(int i=col.size()-1; i>=0; i--){
            int curr=col[i]-'A'+1;
            column+=pow*curr;
            pow*=26;
        }
        cout<<"R"+row+"C"<<to_string(column)<<endl;


    }







}

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    clock_t z = clock();

    int t = 1;
    cin >> t;
    while (t--)
        solve();

    return 0;
}