In c++ given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.

The overall run time complexity should be O(log (m+n)).

Input

In the first line enter the size of the first array

In the second line enter the list of numbers of the first array

In the third line enter the size of the second array

In the last line enter the list of numbers of the second array

Output

The output is the median of the two sorted arrays

Example of input

2 1 3 1 2 

Example of output

The median is 2 

Constraints:

nums1.length == m

nums2.length == n

0 <= m <= 1000

0 <= n <= 1000

1 <= m + n <= 2000

-106 <= nums1[i], nums2[i] <= 106

GIVEN CODE TO WORK WITH:

#include #include #include

using namespace std;

double findMedianSortedArrays(vector& nums1, vector& nums2) { //********************************* //your code here //*********************************** }

int main(){ vector firstList; vector secondList; int sizeFirstList; int sizeSecondList; int numbersList1; int numbersList2;

//Enter size of vector 1: cin>>sizeFirstList; if(sizeFirstList<0 || sizefirstlist>1000) { cout<<"Try again and enter an integer greater than or equal to 0, and less than or equal to 1000"<> numbersList1; if(numbersList1<-106 || numbersList1>106){ cout<<"Try again and enter an integer greater than or equal to -106, and less than or equal to 106"<

//Enter size of vector 2: cin>>sizeSecondList; if(sizeSecondList<0 || sizesecondlist>1000) { cout<<"Try again and enter an integer greater than or equal to 0, and less than or equal to 1000"<> numbersList2; if(numbersList2<-106 || numbersList2>106){ cout<<"Try again and enter an integer greater than or equal to -106, and less than or equal to 106"<

cout<<"The median is "<