description: Alice has just learned addition. However, she hasn't learned the concept of "carrying" fully — instead of carrying to the next column, she carries to the column two columns to the left.
  
  For example, the regular way to evaluate the sum 2039 + 2976 would be as shown: 
  
  <image>
  
  However, Alice evaluates it as shown: 
  
  <image>
  
  In particular, this is what she does: 
  
    * add 9 and 6 to make 15, and carry the 1 to the column two columns to the left, i. e. to the column "0 9"; 
    * add 3 and 7 to make 10 and carry the 1 to the column two columns to the left, i. e. to the column "2 2"; 
    * add 1, 0, and 9 to make 10 and carry the 1 to the column two columns to the left, i. e. to the column above the plus sign; 
    * add 1, 2 and 2 to make 5; 
    * add 1 to make 1. 
  
  Thus, she ends up with the incorrect result of 15005.
  
  Alice comes up to Bob and says that she has added two numbers to get a result of n. However, Bob knows that Alice adds in her own way. Help Bob find the number of ordered pairs of positive integers such that when Alice adds them, she will get a result of n. Note that pairs (a, b) and (b, a) are considered different if a ≠ b.
  
  Input
  
  The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
  
  The only line of each test case contains an integer n (2 ≤ n ≤ 10^9) — the number Alice shows Bob.
  
  Output
  
  For each test case, output one integer — the number of ordered pairs of positive integers such that when Alice adds them, she will get a result of n. 
  
  Example
  
  Input
  
  
  5
  100
  12
  8
  2021
  10000
  
  
  Output
  
  
  9
  4
  7
  44
  99
  
  Note
  
  In the first test case, when Alice evaluates any of the sums 1 + 9, 2 + 8, 3 + 7, 4 + 6, 5 + 5, 6 + 4, 7 + 3, 8 + 2, or 9 + 1, she will get a result of 100. The picture below shows how Alice evaluates 6 + 4: 
  
  <image>
solution: #include <bits/stdc++.h>
  using namespace std;
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int f = 0;
    cerr << '{';
    for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  const int N = 2e5 + 1;
  long long fact[N], inv[N], invfact[N];
  vector<int> primes, lpf;
  void sieve(int n) {
    primes.assign(1, 2);
    lpf.assign(n + 1, 2);
    for (int x = 3; x <= n; x += 2) {
      if (lpf[x] == 2) primes.push_back(lpf[x] = x);
      for (int i = 0;
           i < primes.size() && primes[i] <= lpf[x] && primes[i] * x <= n; ++i)
        lpf[primes[i] * x] = primes[i];
    }
  }
  long long mul(long long a, long long b, const long long MOD = 1000000007) {
    long long res = 0;
    for (a %= MOD; b > 0; a <<= 1, b >>= 1) {
      if (a >= MOD) a -= MOD;
      if (b & 1) {
        res += a;
        if (res >= MOD) res -= MOD;
      }
    }
    return res;
  }
  long long power(long long x, long long n, const long long MOD = 1000000007) {
    long long res = 1;
    for (; n > 0; n >>= 1) {
      if (n & 1) res = mul(res, x, MOD);
      x = mul(x, x, MOD);
    }
    return res;
  }
  bool mr_test(long long n, int a, long long d, int s) {
    long long x = power(a, d, n);
    if (x == 1 || x == n - 1) return false;
    for (int r = 1; r < s; ++r) {
      x = mul(x, x, n);
      if (x == n - 1) return false;
    }
    return true;
  }
  bool miller_rabin(long long n, int k = 5) {
    if (n < 4) return n == 2 || n == 3;
    if (n % 2 == 0 || n % 3 == 0) return false;
    int s = __builtin_ctz(n - 1);
    long long d = (n - 1) >> s;
    for (int it = 1; it <= 5; ++it) {
      int a = 2 + rand() % (n - 3);
      if (mr_test(n, a, d, s)) return false;
    }
    return true;
  }
  bool isPrime(long long n) {
    if (n < 2) return false;
    if (n < lpf.size()) return lpf[n] == n;
    return miller_rabin(n);
  }
  long long div(long long n) {
    long long ans = 1;
    for (int i = 0; i < primes.size(); i++) {
      if (primes[i] * primes[i] * primes[i] > n) break;
      long long cnt = 1;
      while (n % primes[i] == 0) {
        n /= primes[i];
        cnt++;
      }
      ans = ans * cnt;
    }
    long long x = round(sqrt(n));
    if (isPrime(n))
      ans *= 2;
    else if (x * x == n) {
      if (isPrime(x)) ans *= 3;
    } else
      ans *= 4;
    return ans;
  }
  int log_(int a, int b) { return (a > b - 1) ? 1 + log_(a / b, b) : 0; }
  long long add(long long a, long long b, long long Mod = 1000000007) {
    if ((a += b) >= Mod)
      a -= Mod;
    else if (a < 0)
      a += Mod;
    return a;
  }
  long long nCr(int n, int r) {
    if (r > n) return 0;
    return mul(mul(fact[n], invfact[r]), invfact[n - r]);
  }
  long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
  void factInverse() {
    fact[0] = inv[1] = fact[1] = invfact[0] = invfact[1] = 1;
    for (long long i = 2; i < N; i++) {
      fact[i] = (fact[i - 1] * i) % 1000000007;
      inv[i] = 1000000007 - (inv[1000000007 % i] * (1000000007 / i) % 1000000007);
      invfact[i] = (inv[i] * invfact[i - 1]) % 1000000007;
    }
  }
  void solve(int K) {
    long long n, m, q, z, k;
    long long ans = 0, cnt = 0;
    long long a, b, c;
    long long x, y;
    long long l{-1}, r{-1};
    string s;
    string t;
    cin >> n;
    if (n < 10) {
      cout << n - 1 << "\n";
      return;
    }
    bitset<11> bit;
    s = to_string(n);
    m = s.size();
    long long tmp = 1;
    y = bit.to_ullong();
    z = 1LL << (m - 2);
    reverse((s).begin(), (s).end());
    while (1) {
      tmp = 1;
      ;
      if (y == (z)) break;
      for (int i = m - 2; i >= 0; i -= 2) {
        if (i == 0 || i == 1) {
          if (bit[i]) {
            tmp *= (9 - (s[i] - '0'));
          } else {
            tmp *= (s[i] - '0' + 1);
          }
          break;
        }
        if (bit[i - 2]) {
          if (bit[i]) {
            tmp *= (9 - (s[i] - '0') + 1);
          } else
            tmp *= (s[i] - '0');
        } else {
          if (bit[i]) {
            tmp *= (9 - (s[i] - '0'));
          } else
            tmp *= (s[i] - '0' + 1);
        }
      }
      for (int i = m - 1; i >= 0; i -= 2) {
        if (i == 0 || i == 1) {
          if (bit[i]) {
            tmp *= (9 - (s[i] - '0'));
          } else {
            tmp *= (s[i] - '0' + 1);
          }
          break;
        }
        if (bit[i - 2]) {
          if (bit[i]) {
            tmp *= (9 - (s[i] - '0') + 1);
          } else
            tmp *= (s[i] - '0');
        } else {
          if (bit[i]) {
            tmp *= (9 - (s[i] - '0'));
          } else
            tmp *= (s[i] - '0' + 1);
        };
      };
      ans += tmp;
      y++;
      bit = y;
    };
    cout << ans - 2 << "\n";
    return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t, k;
    k = 1;
    t = 1;
    cin >> t;
    while (t--) {
      solve(k);
      k++;
    }
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
    cout << flush;
    return 0;
  }
  