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
#include <bits/stdc++.h>
auto clk = clock();
const long double PI2 = acos(-1) / 2.0l;
const int dx[] = {0, 0, -1, 1}, dy[] = {-1, 1, 0, 0};
const int dx8[] = {-1, -1, -1, 0, 1, 1, 1, 0},
          dy8[] = {-1, 0, 1, 1, 1, 0, -1, -1};
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MAXN = 100010;
long long Gcd(long long a, long long b) {
  if (a % b == 0) return b;
  return Gcd(b, a % b);
}
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long Min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
void printvector(std::vector<long long> v) {
  for (int i = 0; i < v.size(); ++i) {
    cout << v[i] << " ";
  }
  cout << "" << endl;
}
void printarray(long long a[], long long n) {
  for (long long i = 0; i < n; ++i) {
    cout << a[i] << " ";
  }
  cout << "" << endl;
}
long long findlcm(vector<long long> arr, long long n) {
  long long ans = arr[0];
  for (int i = 1; i < n; i++) ans = (((arr[i] * ans)) / (Gcd(arr[i], ans)));
  return ans;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long lowerbound(long long k, vector<long long> v) {
  long long l = 0;
  long long size = v.size();
  long long r = size - 1;
  long long mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (v[mid] < k)
      l = mid + 1;
    else {
      r = mid - 1;
    }
  }
  return l;
}
long long upperbound(long long k, vector<long long> v) {
  long long l = 0;
  long long size = v.size();
  long long r = size - 1;
  long long mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (v[mid] > k)
      r = mid - 1;
    else {
      l = mid + 1;
    }
  }
  return l;
}
long long spf[MAXN];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < MAXN; i++) spf[i] = i;
  for (int i = 4; i < MAXN; i += 2) spf[i] = 2;
  for (int i = 3; i * i < MAXN; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < MAXN; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
long long n, m;
vector<long long> shortPath(vector<pair<long long, long long>> v[],
                            long long src, long long V = 0) {
  set<pair<long long, long long>> setds;
  long long sizee = V;
  long long IN = 1000000000000000000;
  vector<long long> dist(sizee, IN);
  vector<bool> vis(sizee, false);
  setds.insert(make_pair(0, src));
  dist[src] = 0;
  while (!setds.empty()) {
    pair<long long, long long> tmp = *(setds.begin());
    setds.erase(setds.begin());
    long long u = tmp.second;
    if (vis[u]) continue;
    vis[u] = true;
    vector<pair<long long, long long>>::iterator i;
    for (auto i = v[u].begin(); i != v[u].end(); ++i) {
      long long v = (*i).first;
      long long weight = (*i).second;
      if (dist[v] > dist[u] + weight) {
        if (dist[v] != 1000000000000000000)
          setds.erase(setds.find(make_pair(dist[v], v)));
        dist[v] = dist[u] + weight;
        setds.insert(make_pair(dist[v], v));
      }
    }
  }
  return dist;
}
long long root(long long a[], long long r) {
  if (a[r] == r) {
    return r;
  } else {
    long long x = root(a, a[r]);
    a[r] = x;
    return x;
  }
}
void unio(long long a[], long long size[], long long x, long long y) {
  x = root(a, x);
  y = root(a, y);
  if (x == y) {
    return;
  }
  if (size[x] >= size[y]) {
    a[y] = x;
    size[x] += size[y];
  } else {
    a[x] = y;
    size[y] += size[x];
  }
}
void CV(long long txt) {
  string s;
  cin >> s;
  map<char, long long> ma;
  ma['0'] = 0;
  ma['1'] = 1;
  ma['2'] = 2;
  ma['3'] = 3;
  ma['4'] = 4;
  ma['5'] = 5;
  ma['6'] = 6;
  ma['7'] = 7;
  ma['8'] = 8;
  ma['9'] = 9;
  long long x = 10;
  for (long long i = 0; i < 26; i++) {
    char c = 'A' + i;
    ma[c] = x++;
  }
  for (long long i = 0; i < 26; i++) {
    char c = 'a' + i;
    ma[c] = x++;
  }
  ma['-'] = 62;
  ma['_'] = 63;
  long long r = 1;
  long long po = 0;
  long long res = 1;
  for (long long i = s.length() - 1; i >= 0; i--) {
    long long c = 0;
    po = ma[s[i]];
    for (int j = 0, _n = (6); j < _n; j++) {
      if ((po & (1 << j)) == 0) c++;
    }
    res = (power(3, c, 1000000007) * res) % 1000000007;
    po = (po + (r * ma[s[i]]) % 1000000007) % 1000000007;
    r = (r * 64) % 1000000007;
  }
  {
    cout << res << endl;
    cerr << endl;
    return;
  };
  vector<long long> bin;
  {
    cout << res << endl;
    cerr << endl;
    return;
  };
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long x68 = 1;
  for (long long vi = 1; vi <= x68; vi++) {
    CV(vi);
  }
  return 0;
}
