Problem Statement

Dr. Suposupo developed a programming language called Shipura. Shipura supports only one binary operator ${\tt >>}$ and only one unary function ${\tt S<\ >}$.

$x {\tt >>} y$ is evaluated to $\lfloor x / 2^y \rfloor$ (that is, the greatest integer not exceeding $x / 2^y$), and ${\tt S<} x {\tt >}$ is evaluated to $x^2 \bmod 1{,}000{,}000{,}007$ (that is, the remainder when $x^2$ is divided by $1{,}000{,}000{,}007$).

The operator ${\tt >>}$ is left-associative. For example, the expression $x {\tt >>} y {\tt >>} z$ is interpreted as $(x {\tt >>} y) {\tt >>} z$, not as $x {\tt >>} (y {\tt >>} z)$. Note that these parentheses do not appear in actual Shipura expressions.

The syntax of Shipura is given (in BNF; Backus-Naur Form) as follows:


expr   ::= term | expr sp ">>" sp term
term   ::= number | "S" sp "<" sp expr sp ">"
sp     ::= "" | sp " "
number ::= digit | number digit
digit  ::= "0" | "1" | "2" | "3" | "4" | "5" | "6" | "7" | "8" | "9"

The start symbol of this syntax is $\tt expr$ that represents an expression in Shipura. In addition, $\tt number$ is an integer between $0$ and $1{,}000{,}000{,}000$ inclusive, written without extra leading zeros.

Write a program to evaluate Shipura expressions.

Input

The input is a sequence of datasets. Each dataset is represented by a line which contains a valid expression in Shipura.

A line containing a single ${\tt \\#}$ indicates the end of the input. You can assume the number of datasets is at most $100$ and the total size of the input file does not exceed $2{,}000{,}000$ bytes.

Output

For each dataset, output a line containing the evaluated value of the expression.

Sample Input


S< S< 12 >> 2 > >
123 >> 1 >> 1
1000000000   >>129
S<S<S<S<S<2>>>>>
S  <S< S<2013    >>> 11 >>> 10 >


Output for the Sample Input


81
30
0
294967268
14592400





Example

Input

S< S< 12 >> 2 > >
123 >> 1 >> 1
1000000000   >>129
S<S<S<S<S<2>>>>>
S  <S< S<2013    >>> 11 >>> 10 >
#


Output

81
30
0
294967268
14592400
#include <bits/stdc++.h>
#define REP(i,n) for(int i=0; i<(int)(n); ++i)

using namespace std;

struct Parser{
    typedef long long LL;
    static const LL MOD = 1000000007;
    struct Result{
        int idx;
        LL val;
        Result(int i, LL v) : idx(i), val(v) {}
    };
    string s;
    Parser(string s_) {
        stringstream ss(s_);
        string t;
        while(ss >> t){
            s += t;
        }
    }
    inline void inc(int& k, char c) const {
        if(s[k] != c){
            printf("assert s[%d] = %c but %c\n", k, c, s[k]);
            exit(1);
        }
        k++;
    }
    inline bool is_term(char c) const {
        if(isdigit(c)) return true;
        return c == 'S';
    }
    int parse() const {
        return expr(0).val;
    }
    Result expr(int k) const {
        // printf("into expr(%d)\n", k);
        Result r = term(k);
        while(s[r.idx] == '>' && r.idx + 2 < s.size() && is_term(s[r.idx + 2])){
            inc(r.idx, '>');
            inc(r.idx, '>');
            // printf("expr(%d) call term(%d)\n", k, r.idx);
            Result t = term(r.idx);
            // printf("back expr(%d)\n", k);
            r.val = r.val >> min(60LL, t.val);
            r.idx = t.idx;
        }
        return r;
    }
    Result term(int k) const {
        // int fk = k;
        // printf("into term(%d)\n", k);
        if(s[k] == 'S'){
            inc(k, 'S');
            inc(k, '<');
            // printf("term(%d) call expr(%d)\n", fk, k);
            Result r = expr(k);
            // printf("back term(%d) (%d, %d)\n", fk);
            r.val = r.val * r.val % MOD;
            inc(r.idx, '>');
            return r;
        }else{
            LL num = 0;
            while(isdigit(s[k])){
                num = num * 10 + s[k++] - '0';
            }
            return Result(k, num);
        }
    }
};

int main(){
    string s;
    while(getline(cin, s) && s != "#"){
        cout << Parser(s).parse() << endl;
    }
    return 0;
}