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
//------------------------------------------
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <bits/stdc++.h>
#include<queue>
#include <numeric>
using namespace std;
 
//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
 
//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}
 template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
 
//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
 
//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define fore(i,a) for(auto &i:a)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
 
//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))
 
//debug
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
 

LL gcd(LL a, LL b){
    if(b == 0) return a;
    return gcd(b,a%b);
}

LL lcm(LL a, LL b){
    LL g = gcd(a,b);
    return a/g *b;
}

int main(){
    int N;
    cin >> N;
    long long A[N],B[N];
    REP(i,N) cin >> A[i];
    REP(i,N) cin >> B[i];
    long long count=0;
    long long ans=0;
    long long plus_a=0,plus_b=0;
    for(int i=0;i<N;++i){
        if(A[i]<B[i]){
            plus_a+=(B[i]-A[i]+1)/2;
        }
        else{
            plus_b+=A[i]-B[i];
        }
}
long long sousa=0;
for(int i=0;i<N;++i){
    sousa+=B[i]-A[i];
}
ans=max(plus_b,plus_a);
if(ans<=sousa){
    cout << "Yes";
}
else{
    cout << "No";
}
}