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>
using namespace std;

#define REP(i,a,b) for(int i=a;i<(int)b;i++)
#define rep(i,n) REP(i,0,n)

typedef long long ll;

int const MOD = 1e9+7;

string line; string::iterator iter;
inline void debug() { cout << "Remain: \"" <<string(iter, line.end())<<'"' << endl; }

void consume(char e) {
  if(*iter == e) { iter++; }
  else { assert(0); }
}

bool ifconsume(char e) {
  if(*iter == e) { consume(e); return 1; }
  return 0;
}

bool strcons(string s) {
  string::iterator is = s.begin();
  string::iterator tmpiter = iter;
  
  while(is!=s.end() &&
        tmpiter!=line.end() &&
        *is == *tmpiter) { tmpiter++; is++; }

  if(is == s.end()) {
    iter = tmpiter;
    return true;
  }
  return false;
}

ll get_num() {
  ll ret = 0;
  while(isdigit(*iter)) { ret *= 10; ret += (*iter)-'0'; consume(*iter); }
  return ret;
}

ll expr();
ll term() {
  if(isdigit(*iter)) { return get_num(); }
  else if(ifconsume('S')) {
    consume('<');
    ll r = expr(); (r*=r) %= MOD;
    consume('>');
    return r;
  }
  else {
    debug();assert(0&&"term");
  }
}

ll expr() {
  ll t1 = term();
  while(strcons(">>")) {
    if(*iter != 'S' && !isdigit(*iter)) { iter -= 2; break; }
    int k = term();
    while(k--) { t1 >>= 1; if(t1 == 0) break; } // log テ・ツ崢榲」ツ?ァテ」ツ?凖」ツ?ケテ」ツ?ヲテ」ツ?ョテヲツ閉エテヲツ閉ーテ」ツつ津・ツーツステ」ツ?湘」ツ?崚」ツつ凝」ツ?ョテ」ツ?ァテ」ツ??ゥツ?氾、ツクツュテ」ツ?ァ == 0 テ」ツ?ョテヲツ楪敕・ツ按暗」ツつ甘」ツつ津・ツ?・テ」ツつ古」ツつ凝」ツ??」ツ?妥」ツ?ァテゥツ?堙」ツつ凝」ツつ暗」ツ??」ツ?ォテ」ツ?ェテ」ツつ凝」ツ??
  }
  return t1;
}

int main() {
  for(string s;getline(cin, s);) {
    if(s == "#") break;
    line.clear();
    int sz=s.size(); rep(i,sz) if(s[i]!=' ') { line.push_back(s[i]); }
    iter = line.begin();
    cout << expr() << endl;
  }
  
  return 0;
}