B. AND 0, Sum Big-Codeforces Round #716 (Div. 2)

Problem - 1514B - Codeforces

B. AND 0, Sum Big

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Baby Badawy's first words were "AND 0 SUM BIG", so he decided to solve the following problem. Given two integers nn and kk, count the number of arrays of length nn such that:

• all its elements are integers between 00 and 2k−12k−1 (inclusive);
• the bitwise AND of all its elements is 00;
• the sum of its elements is as large as possible.

Since the answer can be very large, print its remainder when divided by 109+7109+7.

Input

The first line contains an integer tt (1≤t≤101≤t≤10) — the number of test cases you need to solve.

Each test case consists of a line containing two integers nn and kk (1≤n≤1051≤n≤105, 1≤k≤201≤k≤20).

Output

For each test case, print the number of arrays satisfying the conditions. Since the answer can be very large, print its remainder when divided by 109+7109+7.

Example

input

Copy

2
2 2
100000 20

output

Copy

4
226732710

Note

In the first example, the 44 arrays are:

• [3,0][3,0],
• [0,3][0,3],
• [1,2][1,2],
• [2,1][2,1].

==================================================================================================================================================

&运算要求只要有一个是0结果一定是0，而我们如果想让结果最大，最好的方法当然是n个2^k-1,但这样与运算之后并不是0,。我们可以让n-1个数字全是2^k-1,剩下一个0，也就是k位都是0的情况。然而这k个位置是0的情况随机分配在n个数字是并不影响答案的大小的，故答案方案数就是n^k，快速幂解决即可

# include<iostream>
# define mod 1000000007
using namespace std;
 
typedef long long int ll;
 
 
ll quickpow(ll base, ll pow)
{
 
    ll ans=1;
 
    while(pow)
    {
 
        if(pow&1)
            ans=ans*base%mod;
 
 
        pow>>=1;
 
        base=base*base%mod;
 
 
    }
 
    return ans;
 
}
int main ()
{
 
      int t;
 
      cin>>t;
 
 
      while(t--)
      {
          ll n,k;
 
          cin>>n>>k;
 
 
          cout<<quickpow(n,k)<<endl;
 
      }
    return 0;
 
}