The answer must be in WinMIPS64 code. Please don't respond with any other code.

Write a WinMIPS64 code to determine whether a given number is an Armstrong number. The n-digit numbers equal to sum of nth powers of their digits (a finite sequence), called Armstrong numbers. They first few are given by 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748,.

For example, consider 371, 371 = 3³ + 7³+ 1³ = 27 + 343 + 1 = 371. A sample C code for your assistance is given below:

#include

#include

int main(){

int num,i;

int digit,sum,capture;

sum = 0.0;

capture = 0.0;

printf( "Enter the total number of digits " );

scanf( "%d", &num );

for( i = 0; i < num; i++ ) {

printf( "Enter the %d digit: ", i+1 );

scanf( "%d", &digit );

sum = sum + pow(digit,num);

capture = capture * 10 + digit;

}

printf( "Entered Number: %d ", capture );

printf( "Sum of Product: %d ", sum );

if( capture == sum )

printf( "Armstrong number" );

else

printf( "Not an Armstrong number" );

}

All examples can be used for manipulation. Note that the numbers displayed are in floating point. Display the numbers using double format in WinMIPS64. You need to test the code only for 1-4 digit numbers from 0 to 9999. It is sufficient to follow the mentioned C logic that checks for 4-digit Armstrong numbers, even though more digits can be specified. Provide a complete report of your code i.e. stall count, code size, CPI.