#include <bits/stdc++.h>
using namespace std;
map<char, int> mp;
string tobinary(int rkm) {
  stack<char> st;
  while (rkm > 0) {
    if ((rkm & 1) == 0)
      st.push('0');
    else
      st.push('1');
    rkm /= 2;
  }
  string s = "";
  int temp = 6 - st.size();
  while (!st.empty()) {
    s += st.top();
    st.pop();
  }
  string ss = "";
  for (int i = 0; i < temp; ++i) ss += '0';
  return ss + s;
}
long long getans(int rkm) {
  string binary = tobinary(rkm);
  int ans = 1;
  for (int i = 0; i < binary.size(); ++i) {
    if (binary[i] == '0') ans *= 3;
  }
  return (ans) % 1000000007;
}
int main() {
  long long ans = 1;
  bool flag1 = false;
  bool flag2 = false;
  bool flag3 = false;
  int j = 0;
  char c1 = '0';
  char c2 = 'A';
  char c3 = 'a';
  for (int i = 0; i < 62; ++i, ++j) {
    if (i == 0) {
      flag1 = true;
      flag2 = false;
      flag3 = false;
      j = 0;
    }
    if (i == 10) {
      flag2 = true;
      flag1 = false;
      flag3 = false;
      j = 0;
    }
    if (i == 36) {
      flag3 = true;
      flag1 = false;
      flag2 = false;
      j = 0;
    }
    if (flag1) mp[c1 + j] = i;
    if (flag2) mp[c2 + j] = i;
    if (flag3) mp[c3 + j] = i;
  }
  mp['-'] = 62;
  mp['_'] = 63;
  string s, ss = "";
  cin >> s;
  for (int i = 0; i < s.size(); ++i) {
    int rkm = getans(mp[s[i]]);
    ans *= rkm;
    ans %= 1000000007;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
map<char, int> mp;
string tobinary(int rkm) {
  stack<char> st;
  while (rkm > 0) {
    if ((rkm & 1) == 0)
      st.push('0');
    else
      st.push('1');
    rkm /= 2;
  }
  string s = "";
  int temp = 6 - st.size();
  while (!st.empty()) {
    s += st.top();
    st.pop();
  }
  string ss = "";
  for (int i = 0; i < temp; ++i) ss += '0';
  return ss + s;
}
long long getans(int rkm) {
  string binary = tobinary(rkm);
  int ans = 1;
  for (int i = 0; i < binary.size(); ++i) {
    if (binary[i] == '0') ans *= 3;
  }
  return (ans) % 1000000007;
}
int main() {
  long long ans = 1;
  bool flag1 = false;
  bool flag2 = false;
  bool flag3 = false;
  int j = 0;
  char c1 = '0';
  char c2 = 'A';
  char c3 = 'a';
  for (int i = 0; i < 62; ++i, ++j) {
    if (i == 0) {
      flag1 = true;
      flag2 = false;
      flag3 = false;
      j = 0;
    }
    if (i == 10) {
      flag2 = true;
      flag1 = false;
      flag3 = false;
      j = 0;
    }
    if (i == 36) {
      flag3 = true;
      flag1 = false;
      flag2 = false;
      j = 0;
    }
    if (flag1) mp[c1 + j] = i;
    if (flag2) mp[c2 + j] = i;
    if (flag3) mp[c3 + j] = i;
  }
  mp['-'] = 62;
  mp['_'] = 63;
  string s, ss = "";
  cin >> s;
  for (int i = 0; i < s.size(); ++i) {
    int rkm = getans(mp[s[i]]);
    ans *= rkm;
    ans %= 1000000007;
  }
  printf("%lld\n", ans);
  return 0;
}

```