#include <bits/stdc++.h>
using namespace std;
const int M = 100005;
long long dp[M];
vector<char> v;
int to(char c) {
  if (c >= '0' && c <= '9') return c - '0';
  if (c >= 'A' && c <= 'Z')
    return c - 'A' + 10;
  else if (c >= 'a' && c <= 'z')
    return c - 'a' + 36;
  else if (c == '-')
    return 62;
  else
    return 63;
}
char from(int x) {
  if (x >= 0 && x <= 9) return x + '0';
  if (x >= 10 && x <= 35)
    return x - 10 + 'A';
  else if (x >= 36 && x <= 61)
    return x - 36 + 'a';
  else if (x == 62)
    return '-';
  else
    return '_';
}
string second;
vector<int> t;
map<char, int> m;
int main() {
  cin >> second;
  for (int i = 0; i < 10; i++) v.push_back('0' + i);
  for (int i = 0; i < 26; i++) v.push_back('A' + i);
  for (int i = 0; i < 26; i++) v.push_back('a' + i);
  v.push_back('-');
  v.push_back('_');
  for (int i = 0; i < v.size(); i++) {
    int x = 0;
    for (int j = 0; j < v.size(); j++) {
      for (int k = 0; k < v.size(); k++) {
        int a = to(v[j]);
        int b = to(v[k]);
        if (from(a & b) == v[i]) {
          x++;
        }
      }
    }
    t.push_back(x);
  }
  for (int i = 0; i < t.size(); i++) m[v[i]] = t[i];
  dp[0] = 1;
  for (int i = 1; i <= second.size(); i++) {
    dp[i] = (dp[i] + 1LL * m[second[i - 1]] * dp[i - 1]) % 1000000007;
  }
  cout << dp[second.size()];
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

While walking down the street Vanya saw a label "Hide&Seek". Because he is a programmer, he used & as a bitwise AND for these two words represented as a integers in base 64 and got new word. Now Vanya thinks of some string s and wants to know the number of pairs of words of length |s| (length of s), such that their bitwise AND is equal to s. As this number can be large, output it modulo 109 + 7.

To represent the string as a number in numeral system with base 64 Vanya uses the following rules:

  * digits from '0' to '9' correspond to integers from 0 to 9; 
  * letters from 'A' to 'Z' correspond to integers from 10 to 35; 
  * letters from 'a' to 'z' correspond to integers from 36 to 61; 
  * letter '-' correspond to integer 62; 
  * letter '_' correspond to integer 63. 

Input

The only line of the input contains a single word s (1 ≤ |s| ≤ 100 000), consisting of digits, lowercase and uppercase English letters, characters '-' and '_'.

Output

Print a single integer — the number of possible pairs of words, such that their bitwise AND is equal to string s modulo 109 + 7.

Examples

Input

z


Output

3


Input

V_V


Output

9


Input

Codeforces


Output

130653412

Note

For a detailed definition of bitwise AND we recommend to take a look in the corresponding article in Wikipedia.

In the first sample, there are 3 possible solutions: 

  1. z&_ = 61&63 = 61 = z
  2. _&z = 63&61 = 61 = z
  3. z&z = 61&61 = 61 = z

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int M = 100005;
long long dp[M];
vector<char> v;
int to(char c) {
  if (c >= '0' && c <= '9') return c - '0';
  if (c >= 'A' && c <= 'Z')
    return c - 'A' + 10;
  else if (c >= 'a' && c <= 'z')
    return c - 'a' + 36;
  else if (c == '-')
    return 62;
  else
    return 63;
}
char from(int x) {
  if (x >= 0 && x <= 9) return x + '0';
  if (x >= 10 && x <= 35)
    return x - 10 + 'A';
  else if (x >= 36 && x <= 61)
    return x - 36 + 'a';
  else if (x == 62)
    return '-';
  else
    return '_';
}
string second;
vector<int> t;
map<char, int> m;
int main() {
  cin >> second;
  for (int i = 0; i < 10; i++) v.push_back('0' + i);
  for (int i = 0; i < 26; i++) v.push_back('A' + i);
  for (int i = 0; i < 26; i++) v.push_back('a' + i);
  v.push_back('-');
  v.push_back('_');
  for (int i = 0; i < v.size(); i++) {
    int x = 0;
    for (int j = 0; j < v.size(); j++) {
      for (int k = 0; k < v.size(); k++) {
        int a = to(v[j]);
        int b = to(v[k]);
        if (from(a & b) == v[i]) {
          x++;
        }
      }
    }
    t.push_back(x);
  }
  for (int i = 0; i < t.size(); i++) m[v[i]] = t[i];
  dp[0] = 1;
  for (int i = 1; i <= second.size(); i++) {
    dp[i] = (dp[i] + 1LL * m[second[i - 1]] * dp[i - 1]) % 1000000007;
  }
  cout << dp[second.size()];
  return 0;
}

```