Problem: You are given n positive integers a1, a2, ..., an.

For every ai you need to find a positive integer ki such that the decimal notation of 2ki contains the decimal notation of ai as a substring among its last min(100, length(2ki)) digits. Here length(m) is the length of the decimal notation of m.

Note that you don't have to minimize ki. The decimal notations in this problem do not contain leading zeros.

Input

The first line contains a single integer n (1 ≤ n ≤ 2 000) — the number of integers ai.

Each of the next n lines contains a positive integer ai (1 ≤ ai < 1011).

Output

Print n lines. The i-th of them should contain a positive integer ki such that the last min(100, length(2ki)) digits of 2ki contain the decimal notation of ai as a substring. Integers ki must satisfy 1 ≤ ki ≤ 1050.

It can be shown that the answer always exists under the given constraints. If there are multiple answers, print any of them.

Examples

Input

2
8
2


Output

3
1


Input

2
3
4857


Output

5
20

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class t, class u>
void chmax(t& first, u second) {
  if (first < second) first = second;
}
template <class t, class u>
void chmin(t& first, u second) {
  if (second < first) first = second;
}
template <class t>
using vc = vector<t>;
template <class t>
using vvc = vc<vc<t>>;
using pi = pair<ll, ll>;
using vi = vc<ll>;
template <class t, class u>
ostream& operator<<(ostream& os, const pair<t, u>& p) {
  return os << "{" << p.first << "," << p.second << "}";
}
template <class t>
ostream& operator<<(ostream& os, const vc<t>& v) {
  os << "{";
  for (auto e : v) os << e << ",";
  return os << "}";
}
using uint = unsigned;
using ull = unsigned long long;
template <class t, size_t n>
ostream& operator<<(ostream& os, const array<t, n>& first) {
  return os << vc<t>(first.begin(), first.end());
}
template <ll i, class T>
void print_tuple(ostream&, const T&) {}
template <ll i, class T, class H, class... Args>
void print_tuple(ostream& os, const T& t) {
  if (i) os << ",";
  os << get<i>(t);
  print_tuple<i + 1, T, Args...>(os, t);
}
template <class... Args>
ostream& operator<<(ostream& os, const tuple<Args...>& t) {
  os << "{";
  print_tuple<0, tuple<Args...>, Args...>(os, t);
  return os << "}";
}
template <class t>
void print(t x, ll suc = 1) {
  cout << x;
  if (suc == 1) cout << "\n";
  if (suc == 2) cout << " ";
}
ll read() {
  ll i;
  cin >> i;
  return i;
}
vi readvi(ll n, ll off = 0) {
  vi v(n);
  for (ll i = ll(0); i < ll(n); i++) v[i] = read() + off;
  return v;
}
template <class T>
void print(const vector<T>& v, ll suc = 1) {
  for (ll i = ll(0); i < ll(v.size()); i++)
    print(v[i], i == ll(v.size()) - 1 ? suc : 2);
}
string readString() {
  string s;
  cin >> s;
  return s;
}
template <class T>
T sq(const T& t) {
  return t * t;
}
void yes(bool ex = true) {
  cout << "Yes"
       << "\n";
  if (ex) exit(0);
}
void no(bool ex = true) {
  cout << "No"
       << "\n";
  if (ex) exit(0);
}
void possible(bool ex = true) {
  cout << "Possible"
       << "\n";
  if (ex) exit(0);
}
void impossible(bool ex = true) {
  cout << "Impossible"
       << "\n";
  if (ex) exit(0);
}
constexpr ll ten(ll n) { return n == 0 ? 1 : ten(n - 1) * 10; }
const ll infLL = LLONG_MAX / 3;
const ll inf = infLL;
ll topbit(signed t) { return t == 0 ? -1 : 31 - __builtin_clz(t); }
ll topbit(ll t) { return t == 0 ? -1 : 63 - __builtin_clzll(t); }
ll botbit(signed first) { return first == 0 ? 32 : __builtin_ctz(first); }
ll botbit(ll first) { return first == 0 ? 64 : __builtin_ctzll(first); }
ll popcount(signed t) { return __builtin_popcount(t); }
ll popcount(ll t) { return __builtin_popcountll(t); }
bool ispow2(ll i) { return i && (i & -i) == i; }
ll mask(ll i) { return (ll(1) << i) - 1; }
bool inc(ll first, ll second, ll c) { return first <= second && second <= c; }
template <class t>
void mkuni(vc<t>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
ll rand_int(ll l, ll r) {
  static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
  return uniform_int_distribution<ll>(l, r)(gen);
}
template <class t>
void myshuffle(vc<t>& first) {
  for (ll i = ll(0); i < ll(ll(first.size())); i++)
    swap(first[i], first[rand_int(0, i)]);
}
template <class t>
ll lwb(const vc<t>& v, const t& first) {
  return lower_bound(v.begin(), v.end(), first) - v.begin();
}
constexpr ll p5(ll n) { return n == 0 ? 1 : p5(n - 1) * 5; }
const ll m = p5(17);
void mult(ll& first, ll second) { first = __int128(first) * second % m; }
ll powm(ll first, ll n) {
  ll res = 1;
  while (n) {
    if (n & 1) mult(res, first);
    mult(first, first);
    n >>= 1;
  }
  return res;
}
ll sub(ll first) {
  vi fs{4};
  for (ll i = ll(0); i < ll(16); i++) fs.push_back(5);
  ll k = 0, step = 1, w = 2, ord = p5(16) * 4;
  for (auto f : fs) {
    ll left = powm(w, ord / f), right = powm(first, ord / f);
    ll cur = 1;
    bool found = false;
    for (ll i = ll(0); i < ll(f); i++) {
      if (cur == right) {
        found = true;
        k += step * i;
        const ll z = powm(w, ord - 1);
        for (ll _ = ll(0); _ < ll(i); _++) mult(first, z);
        break;
      }
      mult(cur, left);
    }
    assert(found);
    step *= f;
    ord /= f;
    w = powm(w, f);
  }
  return k;
}
void slv(ll first) {
  {
    first *= ten(6);
    const ll V = 1 << 17;
    first = (first + V - 1) / V * V;
    if (first % 5 == 0) first += V;
    first >>= 17;
  }
  ll k = sub(first) + 17;
  print(k);
  void(0);
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  cout << fixed << setprecision(20);
  if (0)
    while (1) {
      ll n;
      cin >> n;
      cerr << sub(n) << endl;
    }
  ll n;
  cin >> n;
  for (ll _ = ll(0); _ < ll(n); _++) {
    ll first;
    cin >> first;
    slv(first);
  }
}