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 <bits/stdc++.h>
// #pragma GCC optimize ("O3")
// #pragma GCC target ("sse4")
 
using namespace std;
 
#define li long long int
#define ld long double
const int MX1=1e5+5;
const int MX2=1e6+5;
const li inf=1e18;
const li mod=1e9+7;
 
li power(li a,li b){li res=1;while(b){if(b%2)res=(res*a)%mod;
a=(a*a)%mod;b/=2;}return res;}
 
li mmi(li a){return power(a,mod-2);}
 
li cntbit(li n){li cnt=0;while(n){n&=(n-1);cnt++;}return cnt;}

int n;
int a[MX1];
int b[MX1];

void solve(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    li ans=0;
    for(int i=1;i<=n;i++){
        cin>>b[i];
        ans+=1LL*abs(b[i]-a[i]);
    }
    string res="No";
    if(ans%2) res="Yes";
    cout<<res;
}
 
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}