The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:

Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?

Help Jerry Mao solve the long standing problem!

Input

The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.

On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.

Output

For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.

Example

Input


2
2 4


Output


2
4

Note

In first case, for m=2, polynomials that satisfy the constraint are x and 2.

In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
int days[15] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int is_leapyear(int y) { return y % 400 == 0 || (y % 4 == 0 && y % 100 != 0); }
int dx[8] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[8] = {1, -1, 0, 0, -1, 1, -1, 1};
inline long long read() {
  long long x = 0, sign = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') sign = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return sign * x;
}
inline void write(long long x) {
  char r[30];
  int t = 0;
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  do {
    r[t++] = x % 10 + '0';
    x /= 10;
  } while (x);
  while (t) putchar(r[--t]);
}
inline void writesp(long long x) {
  write(x);
  putchar(' ');
}
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
inline void writedo(double x, int op) {
  cout << fixed << setprecision(op) << x << endl;
}
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
long long qmul(long long a, long long b) {
  long long ans = 0;
  while (b) {
    if (b & 1) ans = (ans + a) % mod;
    a = (a + a) % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  int T = read();
  while (T--) {
    long long n = read();
    n += 2;
    long long pos = n / 4 % mod;
    long long ans = (pos * (pos + 1) % mod + (n % 4 >= 2 ? pos + 1 : 0)) % mod;
    writeln(ans);
  }
  return 0;
}
