#include <bits/stdc++.h>
using namespace std;
const long long INF = (long long)LLONG_MAX;
int dx[4] = {0, -1, 0, 1};
int dy[4] = {1, 0, -1, 0};
int dx8[9] = {1, 0, -1, 0, 1, -1, -1, 1, 0};
int dy8[9] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
long long ipow(long long base, long long exp) {
  long long result = 1;
  while (exp) {
    if (exp & 1) result *= base;
    exp >>= 1;
    base *= base;
  }
  return result;
}
string tos(long long n) {
  stringstream ss;
  string ans;
  ss << n;
  ss >> ans;
  return ans;
}
long long toll(string n) { return atoll(n.c_str()); }
double dist(double x1, double y1, double x2, double y2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
}
int doubleCompare(double x, double y) {
  if (fabs(x - y) <= 1e-100) return 0;
  if (x < y) return -1;
  return 1;
}
int main() {
  string x = "";
  for (char i = '0'; i <= '9'; i++) x += i;
  for (char i = 'A'; i <= 'Z'; i++) x += i;
  for (char i = 'a'; i <= 'z'; i++) x += i;
  x += '-';
  x += '_';
  map<char, int> nada;
  map<char, long long> dp;
  for (int i = 0; i < x.size(); i++) nada[x[i]] = i;
  for (int i = 0; i < x.size(); i++) {
    for (int j = 0; j < x.size(); j++) {
      for (int h = 0; h < x.size(); h++) {
        int a = (j & h);
        if (a == nada[x[i]]) dp[x[i]]++;
      }
    }
  }
  string s;
  cin >> s;
  long long ans = 1;
  for (int i = 0; i < s.size(); i++) {
    ans *= dp[s[i]];
    while (ans >= 1e9 + 7) ans -= 1e9 + 7;
    while (ans < 0) ans += 1e9 + 7;
  }
  cout << ans;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 long long INF = (long long)LLONG_MAX;
int dx[4] = {0, -1, 0, 1};
int dy[4] = {1, 0, -1, 0};
int dx8[9] = {1, 0, -1, 0, 1, -1, -1, 1, 0};
int dy8[9] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
long long ipow(long long base, long long exp) {
  long long result = 1;
  while (exp) {
    if (exp & 1) result *= base;
    exp >>= 1;
    base *= base;
  }
  return result;
}
string tos(long long n) {
  stringstream ss;
  string ans;
  ss << n;
  ss >> ans;
  return ans;
}
long long toll(string n) { return atoll(n.c_str()); }
double dist(double x1, double y1, double x2, double y2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
}
int doubleCompare(double x, double y) {
  if (fabs(x - y) <= 1e-100) return 0;
  if (x < y) return -1;
  return 1;
}
int main() {
  string x = "";
  for (char i = '0'; i <= '9'; i++) x += i;
  for (char i = 'A'; i <= 'Z'; i++) x += i;
  for (char i = 'a'; i <= 'z'; i++) x += i;
  x += '-';
  x += '_';
  map<char, int> nada;
  map<char, long long> dp;
  for (int i = 0; i < x.size(); i++) nada[x[i]] = i;
  for (int i = 0; i < x.size(); i++) {
    for (int j = 0; j < x.size(); j++) {
      for (int h = 0; h < x.size(); h++) {
        int a = (j & h);
        if (a == nada[x[i]]) dp[x[i]]++;
      }
    }
  }
  string s;
  cin >> s;
  long long ans = 1;
  for (int i = 0; i < s.size(); i++) {
    ans *= dp[s[i]];
    while (ans >= 1e9 + 7) ans -= 1e9 + 7;
    while (ans < 0) ans += 1e9 + 7;
  }
  cout << ans;
}

```