Problem: We know that lucky digits are digits 4 and 7, however Vasya's got another favorite digit 0 and he assumes it also is lucky! Lucky numbers are such non-negative integers whose decimal record only contains lucky digits. For example, numbers 0, 47, 7074 are lucky, but 1, 7377, 895,  -7 are not.

Vasya has t important positive integers he needs to remember. Vasya is quite superstitious and he wants to remember lucky numbers only, so he is asking you for each important number to represent it as a sum of exactly six lucky numbers (Vasya just can't remember more numbers). Then Vasya can just remember these six numbers and calculate the important number at any moment.

For each of t important integers represent it as the sum of six lucky numbers or state that this is impossible.

Input

The first line contains a single integer t (1 ≤ t ≤ 5000).

Next t lines contain a single positive integer ni (1 ≤ ni ≤ 1018) — the list of important numbers.

Please, do not use the %lld to read or write 64-bit integers С++. It is preferred to read the cin, cout streams or the %I64d specifier.

Output

Print t lines. The i-th line must contain the answer for the i-th important number: if the solution exists, the line must contain exactly six lucky numbers the sum of which equals ni, if the solution doesn't exist the string must contain a single integer -1.

If there are multiple answers print any of them.

Examples

Input

5
42
17
444
7
51


Output

7 7 7 7 7 7
-1
400 0 40 0 4 0
7 0 0 0 0 0
47 4 0 0 0 0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T BigMod(T b, T p, T m) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    T s = BigMod(b, p / 2, m);
    return ((s % m) * (s % m)) % m;
  }
  return ((b % m) * (BigMod(b, p - 1, m) % m)) % m;
}
template <typename T>
T ModInv(T b, T m) {
  return BigMod(b, m - 2, m);
}
template <typename T>
T in() {
  char ch;
  T n = 0;
  bool ng = false;
  while (1) {
    ch = getchar();
    if (ch == '-') {
      ng = true;
      ch = getchar();
      break;
    }
    if (ch >= '0' && ch <= '9') break;
  }
  while (1) {
    n = n * 10 + (ch - '0');
    ch = getchar();
    if (ch < '0' || ch > '9') break;
  }
  return (ng ? -n : n);
}
template <typename T>
T POW(T B, T printf) {
  if (printf == 0) return 1;
  if (printf & 1)
    return B * POW(B, printf - 1);
  else
    return (POW(B, printf / 2) * POW(B, printf / 2));
}
template <typename T>
T Bigmod(T b, T p, T m) {
  if (p == 0)
    return 1;
  else if (!(p & 1))
    return (Bigmod(b, p / 2, m) * Bigmod(b, p / 2, m)) % m;
  else
    return ((b % m) * Bigmod(b, p - 1, m)) % m;
}
template <typename T>
T Dis(T x1, T y1, T x2, T y2) {
  return sqrt((x1 - x2 * x1 - x2) + (y1 - y2 * y1 - y2));
}
template <typename T>
T Angle(T x1, T y1, T x2, T y2) {
  return atan(double(y1 - y2) / double(x1 - x2));
}
template <typename T>
T DIFF(T a, T b) {
  T d = a - b;
  if (d < 0)
    return -d;
  else
    return d;
}
template <typename T>
T ABS(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
template <typename T>
T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
template <typename T>
T euclide(T a, T b, T &x, T &y) {
  if (a < 0) {
    T d = euclide(-a, b, x, y);
    x = -x;
    return d;
  }
  if (b < 0) {
    T d = euclide(a, -b, x, y);
    y = -y;
    return d;
  }
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    T d = euclide(b, a % b, x, y);
    T t = x;
    x = y;
    y = t - (a / b) * y;
    return d;
  }
}
template <typename T>
void ia(T a[], int n) {
  for (int i = 0; i < n; i++) cin >> a[i];
}
template <typename T>
void pa(T a[], int n) {
  for (int i = 0; i < n - 1; i++) cout << a[i] << " ";
  cout << a[n - 1] << endl;
}
template <typename T>
long long int isLeft(T a, T b, T c) {
  return (a.x - b.x) * (b.y - c.y) - (b.x - c.x) * (a.y - b.y);
}
int Set(int N, int pos) { return N = N | (1 << pos); }
int Reset(int N, int pos) { return N = N & ~(1 << pos); }
bool Check(int N, int pos) { return (bool)(N & (1 << pos)); }
template <class T, class first>
inline T togglebit(T a, first i) {
  T t = 1;
  return (a ^ (t << i));
}
int toInt(string s) {
  int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
int toLlint(string s) {
  long long int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
int cdigittoint(char ch) { return ch - '0'; }
bool isVowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
bool isConst(char ch) {
  if (isalpha(ch) && !isVowel(ch)) return true;
  return false;
}
double DEG(double x) { return (180.0 * x) / (2.0 * acos(0.0)); }
double RAD(double x) { return (x * (double)2.0 * acos(0.0)) / (180.0); }
int DP[23][7];
int a[23];
int PT[23][7][2];
int dp(int p, int c) {
  if (p == 0) return c == 0;
  if (DP[p][c] != -1) return DP[p][c];
  int &res = DP[p][c];
  res = 0;
  for (int i = 0; i <= 6; i++) {
    for (int j = 0; j <= 6; j++) {
      int four = i, seven = j;
      if (i + j > 6) continue;
      int dg = (four * 4 + seven * 7 + c) % 10;
      int cr = (four * 4 + seven * 7 + c) / 10;
      if (dg == a[p]) {
        int tmp = dp(p - 1, cr);
        if (tmp) {
          PT[p][c][0] = four;
          PT[p][c][1] = seven;
        }
        res |= tmp;
      }
    }
  }
  return res;
}
long long int ans[8][23];
void Print(int l) {
  int p = l, c = 0;
  memset(ans, 0, sizeof ans);
  while (p) {
    int four = PT[p][c][0], seven = PT[p][c][1];
    for (int i = 1; i <= four; i++) ans[i][p] = 4;
    for (int i = 1; i <= seven; i++) ans[i + four][p] = 7;
    p = p - 1;
    c = (four * 4 + seven * 7 + c) / 10;
  }
  for (int i = 1; i <= 6; i++) {
    int j = 1;
    while (ans[i][j] == 0 && j <= l) j++;
    if (j == l + 1) printf("0");
    while (j <= l) printf("%d", ans[i][j]), j++;
    printf(" ");
  }
  printf("\n");
  ;
}
int main() {
  int(t);
  scanf("%d", &t);
  while (t--) {
    long long int x;
    scanf("%lld", &x);
    int l = 0;
    while (x) {
      a[++l] = x % 10;
      x /= 10;
    }
    reverse(a + 1, a + 1 + l);
    memset(DP, -1, sizeof DP);
    int ans = dp(l, 0);
    if (ans)
      Print(l);
    else
      puts("-1");
  }
  return 0;
}