#include <bits/stdc++.h>
using namespace std;
template <typename A>
inline void read(A& a) {
  a = 0;
  char c = 0;
  int f = 1;
  while (c < '0' || c > '9') (c = getchar()) == '-' ? f = -1 : 0;
  while (c >= '0' && c <= '9') a = (a << 3) + (a << 1) + c - '0', c = getchar();
  f == -1 ? a = -a : 0;
}
char buf[30];
template <typename A>
inline void write(A a) {
  if (a < 0) putchar('-'), a = -a;
  int top = 0;
  if (!a) buf[top = 1] = '0';
  while (a) buf[++top] = a % 10 + '0', a /= 10;
  while (top) putchar(buf[top--]);
}
template <typename A, typename B>
inline bool SMAX(A& x, const B& y) {
  return y > x ? x = y, 1 : 0;
}
template <typename A, typename B>
inline bool SMIN(A& x, const B& y) {
  return y < x ? x = y, 1 : 0;
}
int T, a;
int main() {
  read(T);
  while (T--) {
    read(a);
    int res = 0;
    while (a) res++, a -= ((a) & (-(a)));
    write((1 << res));
    putchar('\n');
  }
}

### Prompt

Please formulate a CPP solution to the following problem:

Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.

Arkadi and Boris Strugatsky. Monday starts on Saturday

Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.

Input

Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.

The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.

Output

For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.

One can show that the number of solutions is always finite.

Example

Input

3
0
2
1073741823


Output

1
2
1073741824

Note

Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:

$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$

For the first value of the parameter, only x = 0 is a solution of the equation.

For the second value of the parameter, solutions are x = 0 and x = 2.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename A>
inline void read(A& a) {
  a = 0;
  char c = 0;
  int f = 1;
  while (c < '0' || c > '9') (c = getchar()) == '-' ? f = -1 : 0;
  while (c >= '0' && c <= '9') a = (a << 3) + (a << 1) + c - '0', c = getchar();
  f == -1 ? a = -a : 0;
}
char buf[30];
template <typename A>
inline void write(A a) {
  if (a < 0) putchar('-'), a = -a;
  int top = 0;
  if (!a) buf[top = 1] = '0';
  while (a) buf[++top] = a % 10 + '0', a /= 10;
  while (top) putchar(buf[top--]);
}
template <typename A, typename B>
inline bool SMAX(A& x, const B& y) {
  return y > x ? x = y, 1 : 0;
}
template <typename A, typename B>
inline bool SMIN(A& x, const B& y) {
  return y < x ? x = y, 1 : 0;
}
int T, a;
int main() {
  read(T);
  while (T--) {
    read(a);
    int res = 0;
    while (a) res++, a -= ((a) & (-(a)));
    write((1 << res));
    putchar('\n');
  }
}

```