Problem: Let's define an unambiguous arithmetic expression (UAE) as follows. 

  * All non-negative integers are UAE's. Integers may have leading zeroes (for example, 0000 and 0010 are considered valid integers). 
  * If X and Y are two UAE's, then "(X) + (Y)", "(X) - (Y)", "(X) * (Y)", and "(X) / (Y)" (all without the double quotes) are UAE's. 
  * If X is an UAE, then " - (X)" and " + (X)" (both without the double quotes) are UAE's.

You are given a string consisting only of digits ("0" - "9") and characters "-", "+", "*", and "/". Your task is to compute the number of different possible unambiguous arithmetic expressions such that if all brackets (characters "(" and ")") of that unambiguous arithmetic expression are removed, it becomes the input string. Since the answer may be very large, print it modulo 1000003 (106 + 3).

Input

The first line is a non-empty string consisting of digits ('0'-'9') and characters '-', '+', '*', and/or '/'. Its length will not exceed 2000. The line doesn't contain any spaces.

Output

Print a single integer representing the number of different unambiguous arithmetic expressions modulo 1000003 (106 + 3) such that if all its brackets are removed, it becomes equal to the input string (character-by-character).

Examples

Input

1+2*3


Output

2


Input

03+-30+40


Output

3


Input

5//4


Output

0


Input

5/0


Output

1


Input

1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1


Output

100728

Note

For the first example, the two possible unambiguous arithmetic expressions are:
    
    
    ((1) + (2)) * (3)  
    (1) + ((2) * (3))

For the second example, the three possible unambiguous arithmetic expressions are:
    
    
    (03) + (( - (30)) + (40))  
    (03) + ( - ((30) + (40)))  
    ((03) + ( - (30))) + (40)

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 2e9;
const long long LINF = 2e16;
const int magic = 348;
const int MOD = 1e6 + 3;
const double eps = 1e-10;
const double pi = acos(-1);
struct fastio {
  static const int S = 1e7;
  char rbuf[S + 48], wbuf[S + 48];
  int rpos, wpos, len;
  fastio() { rpos = len = wpos = 0; }
  inline char Getchar() {
    if (rpos == len) rpos = 0, len = fread(rbuf, 1, S, stdin);
    if (!len) return EOF;
    return rbuf[rpos++];
  }
  template <class T>
  inline void Get(T &first) {
    char ch;
    bool f;
    T res;
    while (!isdigit(ch = Getchar()) && ch != '-') {
    }
    if (ch == '-')
      f = false, res = 0;
    else
      f = true, res = ch - '0';
    while (isdigit(ch = Getchar())) res = res * 10 + ch - '0';
    first = (f ? res : -res);
  }
  inline void getstring(char *s) {
    char ch;
    while ((ch = Getchar()) <= 32) {
    }
    for (; ch > 32; ch = Getchar()) *s++ = ch;
    *s = '\0';
  }
  inline void flush() {
    fwrite(wbuf, 1, wpos, stdout);
    fflush(stdout);
    wpos = 0;
  }
  inline void Writechar(char ch) {
    if (wpos == S) flush();
    wbuf[wpos++] = ch;
  }
  template <class T>
  inline void Print(T first, char ch) {
    char s[20];
    int pt = 0;
    if (first == 0)
      s[++pt] = '0';
    else {
      if (first < 0) Writechar('-'), first = -first;
      while (first) s[++pt] = '0' + first % 10, first /= 10;
    }
    while (pt) Writechar(s[pt--]);
    Writechar(ch);
  }
  inline void printstring(char *s) {
    int pt = 1;
    while (s[pt] != '\0') Writechar(s[pt++]);
  }
} io;
template <typename T>
inline void check_max(T &first, T cmp) {
  first = max(first, cmp);
}
template <typename T>
inline void check_min(T &first, T cmp) {
  first = min(first, cmp);
}
template <typename T>
inline T myabs(T first) {
  return first >= 0 ? first : -first;
}
template <typename T>
inline T gcd(T first, T second) {
  return second == 0 ? first : gcd(second, first % second);
}
inline int add(int first) {
  if (first >= MOD) first -= MOD;
  return first;
}
inline int add(int first, int MO) {
  if (first >= MO) first -= MO;
  return first;
}
inline int sub(int first) {
  if (first < 0) first += MOD;
  return first;
}
inline int sub(int first, int MO) {
  if (first < 0) first += MO;
  return first;
}
inline void Add(int &first, int second) { first = add(first + second); }
inline void Add(int &first, int second, int MO) {
  first = add(first + second, MO);
}
inline void Sub(int &first, int second) { first = sub(first - second); }
inline void Sub(int &first, int second, int MO) {
  first = sub(first - second, MO);
}
template <typename T>
inline int quick_pow(int first, T second) {
  int res = 1;
  while (second) {
    if (second & 1) res = 1ll * res * first % MOD;
    first = 1ll * first * first % MOD;
    second >>= 1;
  }
  return res;
}
template <typename T>
inline int quick_pow(int first, T second, int MO) {
  int res = 1;
  while (second) {
    if (second & 1) res = 1ll * res * first % MO;
    first = 1ll * first * first % MO;
    second >>= 1;
  }
  return res;
}
const int MAXN = 2000;
char s[MAXN + 48];
int len;
int n, type[MAXN + 48];
int dp[MAXN + 48][MAXN + 48], sum[MAXN + 48][MAXN + 48];
int main() {
  io.getstring(s + 1);
  len = strlen(s + 1);
  for (register int i = 1; i <= len; i++) {
    if (isdigit(s[i])) continue;
    if (s[i] == '*' || s[i] == '/') {
      type[++n] = 0;
      if (i == 1 || !isdigit(s[i - 1])) {
        printf("0\n");
        return 0;
      }
    }
    if (s[i] == '+' || s[i] == '-') {
      ++n;
      if (i > 1 && !isdigit(s[i - 1])) type[n] = 1;
    }
  }
  if (!isdigit(s[len])) {
    printf("0\n");
    return 0;
  }
  dp[0][0] = 1;
  sum[0][0] = 1;
  for (register int i = 1; i <= n; i++)
    for (register int j = i; j >= 1; j--) {
      Add(dp[i][j], dp[i - 1][j - 1]);
      if (!type[i]) Add(dp[i][j], sum[i - 1][j]);
      sum[i][j] = add(sum[i][j + 1] + dp[i][j]);
    }
  int ans = 0;
  for (register int i = 0; i <= n; i++) Add(ans, dp[n][i]);
  printf("%d\n", ans);
  io.flush();
  return 0;
}