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: Accepted)

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<stack>
#include<map>
#include<vector>
#include<queue>
#define N 500010
using namespace std;
typedef long long ll;
ll a[N];
ll b[N];
int main()
{
    ll n;
    cin>>n;
    for (ll i=1;i<=n;i++)
       cin>>a[i];
    for (ll i=1;i<=n;i++)
       cin>>b[i];
       ll ans=0;
    for (ll i=1;i<=n;i++)
      {
          if (a[i]>b[i])
               ans-=(a[i]-b[i]);
          else
              ans+=(b[i]-a[i])/2;;
      }
      if (ans>=0)
         cout << "Yes" << endl;
    else
         cout << "No" << endl;
    return 0;
}
/*#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<stack>
#include<map>
#include<vector>
#include<queue>
#define N 500010
using namespace std;
typedef long long ll;
int father[100],Rank[100];
int a[100],b[100];

int Find(int x)   //查找
{
   while (father[x]!=x)
    x=father[x];
   return x;
}
void Union (int x,int y) //合并
{
    int fx,fy;
    fx=Find(x);
    fy=Find(y);
    if (fx!=fy)        //如果x,y不在一个集合里面就合并
        father[fx]=fy;
}
int main()
{
  int n,m,x,w,v;
  cin>>n>>m;
  for (int i=1;i<=n;i++)
    {
        cin>>x;
        father[i]=x;
    }
    for (int i=1;i<=m;i++)
     {
         cin>>w>>v;
         Union(w+1,v+1);
     }
     for (int i=1;i<=n;i++)
     {

     }
    return 0;
}
*/