Problem:
Score : 300 points

Problem Statement

You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N.
Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal.

Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions simultaneously:

Add 2 to a_i.

Add 1 to b_j.

Constraints

1 ≤ N ≤ 10 000

0 ≤ a_i,b_i ≤ 10^9 (1 ≤ i ≤ N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 .. a_N
b_1 b_2 .. b_N

Output

If we can repeat the operation zero or more times so that the sequences a and b become equal, print Yes; otherwise, print No.

Sample Input 1

3
1 2 3
5 2 2

Sample Output 1

Yes

For example, we can perform three operations as follows to do our job:

First operation: i=1 and j=2. Now we have a = \{3,2,3\}, b = \{5,3,2\}.

Second operation: i=1 and j=2. Now we have a = \{5,2,3\}, b = \{5,4,2\}.

Third operation: i=2 and j=3. Now we have a = \{5,4,3\}, b = \{5,4,3\}.

Sample Input 2

5
3 1 4 1 5
2 7 1 8 2

Sample Output 2

No

Sample Input 3

5
2 7 1 8 2
3 1 4 1 5

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<string>
#include<string.h>
#include<map>
#include<set>
#include<vector>
#include<stdio.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int ifarray(int a,int b)
{
int x=0;
while (a<b){
x++;
a+=2;
b++;
}
if (a>b) return 0;
return x;
} 
int main(){
int x;
int a[10001];
int b[10001];
int sma=0;
int smb=0;
cin>>x;
for (int i=1;i<=x;i++)
cin>>a[i];
for (int i=1;i<=x;i++)
cin>>b[i];
for (int i=1;i<=x;i++){
 sma+=a[i];
 smb+=b[i];
}
int ok=1;
for (int i=1;i<=x;i++)
if (a[i]!=b[i])ok=0;
if (sma>smb) {cout<<"No"<<endl;return 0;}
else if (ok) {cout<<"Yes"<<endl;return 0;}
     else if (sma==smb) {cout<<"No"<<endl;return 0;}
int n=ifarray(sma,smb);
if (n) cout<<"";
else {cout<<"No"<<endl;return 0;}
int chaa=0;
int chab=0;
for (int i=1;i<=x;i++)
if (a[i]<=b[i]) chaa+=b[i]-a[i];
           else chab+=a[i]-b[i];
if (chaa/2<chab) {cout<<"No"<<endl;return 0;}
if ((chaa==n*2 && chab==n)||(abs(n*2-chaa==n-chab) && abs(n-chab%2==0))) {cout<<"Yes"<<endl;return 0;}
else {cout<<"No"<<endl;return 0;}
}