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 <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <cctype>
#include <cstdlib>
#include <set>
#include <map>
#include <sstream>
using namespace  std;
#define  ll long long
#define mINF(a) memset(a,INF,sizeof(a))
#define m0(a) memset(a,0,sizeof(a))
#define mf1(a) memset(a,-1,sizeof(a))
#define lson i<<1,l,l+r>>1
#define rson i<<1|1,((l+r)>>1)+1,r
typedef pair<ll,ll>P;
static const ll INF=1e18;
static const ll len=1e4+7;
int main()
{
	ll n,a[len],b[len];
	while(cin>>n)
	{
		for(ll i=0;i<n;++i)
			scanf("%lld",a+i);
		for(ll i=0;i<n;++i)
			scanf("%lld",b+i);
		ll c[len],sum1=0,sum2=0;
		for(ll i=0;i<n;++i)
		{
			sum1+=a[i];
			sum2+=b[i]; 
			c[i]=fabs(a[i]-b[i]);
		}
		if(sum1>sum2)puts("No");
		else
		if(sum1==sum2)
		{
			ll w=0;
			for(ll i=0;i<n;++i)
			{
				if(c[i]!=0)
					w=1;
			}
			if(w==1)puts("No");
			else puts("Yes");
		}
		else
		{
			ll sum3=0;
			for(ll i=0;i<n;++i)
			{
				if(c[i]!=0)
					sum3++;
			}
			if(sum3%2==0)puts("Yes");
			else puts("No");
		}
	}
    return 0;
}