int power (int a, int b) {
  if (b==0)
    return 1;
  if (b==1)
    return a;
  int ans = power(a,b/2);
  if (b%2==0)
    return ans*ans;
  else
    return a*ans*ans;
}

//Binary Search implementation //Time Complexity: O (log n) //return int {-1 , mid index}
int binarySearch(int tar, vector<int>& arr){
int lo=0, md, hi=arr.size()-1;
while(hi>=lo){
md = lo + (hi - lo) / 2;
if(arr[md]==tar) return md;
else if(arr[md]<tar) lo=md+1;
   		else hi=md-1;
}
return -1;
}
---------------------------------------------------------------------------------------------------
//Binary Search by using Stls //Time Complexity: O (log n)
// for binary search in containers like vector (let target element=6)
binary_search(v.begin(), v.end(), 6); 
// return 1 or 0 as present or not
---------------------------------------------------------------------------------------------------
//binary search with lower and upper bound by using stls
int x, ans; //x is passing value, ans is a value you want, v --> vector or any container
ans = lower_bound(v.begin(), v.end(), x) - v.begin(); //ans >= x (equal number or first number after x)
ans = upper_bound(v.begin(), v.end(), x) - v.begin(); // ans > x (first number after x)

//implementation
int lower_bound(vector<int> nums, int target){
        int l = 0, r = nums.size()-1, m = 0;
        while(l < r) {
            m = (l+r)/2;
            if(nums[m] < target)
                l = m+1;
            else 
                r = m;
        }
        return r;
    }

    int upper_bound(vector<int> nums, int target){
        int l = 0, r = nums.size()-1, m = 0;
        while(l < r) {
            m = (l+r)/2;
            if(nums[m] <= target)
                l = m+1;
            else 
                r = m;
        }
        return r;
    }
---------------------------------------------------------------------------------------------------
// Two Pointer implementation
// Two pointer technique based solution to find
// if there is a pair in A[0..N-1] with a given sum.
int isPairSum(int A[], int N, int X)
{
    // represents first pointer
    int i = 0;
  
    // represents second pointer
    int j = N - 1;
  
    while (i < j) {
      
        int curSum = A[i]+arr[j];
      
        // If we find a pair
        if curSum == X)
            return 1;
  
        // If sum of elements at current
        // pointers is less, we move towards
        // higher values by doing i++
        else if (curSum < X)
            i++;
  
        // If sum of elements at current
        // pointers is more, we move towards
        // lower values by doing j--
        else
            j--;
    }
    return 0;
}

//Sum of numbers from 1 ---> n
int n, sum;
sum = (n*(n+1))/2;
 
//flooring integer number (int)        "floor أرضيه"
int a=5, b=2;
int n = a/b; // n= 2
 
//Ceiling integer number (int)          "ceil سقف"
int a=5, b=2;
int ans = (a+(b-1))/b; //n=3
 
//Division Theory 
int a=10, b=3;     // a ---> numerator البسط / b ---> denominator المقام
int ans = a/b;     // n= 3
//meaning:
//1- you can give everyone of b ans
//2- ans is a number of numbers from 1--> a which is divisible by b (find multiple in Range)
 
//Number of Steps

# Infinity, and it's brother minus infinity, 
# comes in handy once in a while.


my_inf = float('Inf') 
99999999 > my_inf 
-> False 
 
my_neg_inf = float('-Inf') 
my_neg_inf > -99999999 
-> False 

def fibonacci(n):
    # return a list of fibonacci numbers
    numbers = [0, 1]
    for i in range(2,n):
        numbers.append(numbers[i-1] + numbers[i-2])
    return numbers[0:n]

def max_by(lst, fn):
  return max(map(fn, lst))

const arrAvg = arr => arr.reduce((a,b) => a + b, 0) / arr.length
// arrAvg([20, 10, 5, 10]) -> 11.25

const arrSum = arr => arr.reduce((a,b) => a + b, 0)
// arrSum([20, 10, 5, 10]) -> 45

const arrMax = arr => Math.max(...arr);
// arrMax([20, 10, 5, 10]) -> 20

const arrMin = arr => Math.min(...arr);
// arrMin([20, 10, 5, 10]) -> 5