solution : Xor Equality | codechef | Problem Code: XOREQUAL

Xor Equality Problem Code: XOREQUAL

For a given N$N$, find the number of ways to choose an integer x$x$ from the range [0,2N−1]$[0,2^N-1]$ such that x⊕(x+1)=(x+2)⊕(x+3)$x\oplus (x+1)=(x+2)\oplus (x+3)$, where ⊕$\oplus$ denotes the bitwise XOR operator.

Since the number of valid x$x$ can be large, output it modulo 109+7${10}^9+7$.

Input

• The first line contains an integer T$T$, the number of test cases. Then the test cases follow.
• The only line of each test case contains a single integer N$N$.

Output

For each test case, output in a single line the answer to the problem modulo 109+7${10}^9+7$.

Constraints

• 1≤T≤105$1\le T\le {10}^5$
• 1≤N≤105$1\le N\le {10}^5$

Subtasks

Subtask #1 (100 points): Original Constraints

Sample Input 1 

2
1
2

Sample Output 1 

1
2

Explanation

Test Case 1$1$: The possible values of x$x$ are {0}$\{0\}$.

Test Case 2$2$: The possible values of x$x$ are {0,2}$\{0,2\}$.

solution : 

#include <bits/stdc++.h>

using namespace std;

#define l 1000000007;

int main() {

int t ;

cin.tie(0);

ios::sync_with_stdio(false);

cin >> t;

while(t--)

{

    long long int n ;

    cin >> n ;

   long  long int count = 1;

    for(int i = 1 ; i <= (n-1); i++ )

    {

        count = (count *2) % l;

    }

    cout << count << "\n";

}

return 0;

}