double root(double n){
// Max and min are used to take into account numbers less than 1
double lo = min(1, n), hi = max(1, n), mid;
// Update the bounds to be off the target by a factor of 10
while(100 * lo * lo < n) lo *= 10;
while(0.01 * hi * hi > n) hi *= 0.1;
for(int i = 0 ; i < 100 ; i++){
mid = (lo+hi)/2;
if(mid*mid == n) return mid;
if(mid*mid > n) hi = mid;
else lo = mid;
}
return mid;
}