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>
 
using namespace std;
 
typedef long long ll;
typedef pair<int, int> P;
static const int INF = 1000010000;
static const int MOD = 1000000007;
 
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define REP(i, n) for(int i = 0; i < (n); ++i)
#define SORT(v) sort(v.begin(), v.end());
#define pb push_back
#define mp make_pair
#define np next_permutation
#define pq priority_queue
  
//int dx4[4] = {0,1,0,-1}, dy4[4] = {-1,0,1,0};
//int dx5[5] = {-1,0,0,0,1}, dy5[5] = {0,-1,0,1,0};
//int dx8[8] = {-1,0,1,1,1,0,-1,-1}, dy8[8] = {1,1,1,0,-1,-1,-1,0};
//int dx9[9] = {-1,0,1,1,1,0,-1,-1,0}, dy9[9] = {1,1,1,0,-1,-1,-1,0,0};

struct edge{
    int to, cost;
};

int main(){
    int n;
    cin >> n;
    int as, bs;
    int a[20010], b[20010];
    REP(i, n){
        cin >> a[i];
    }
    REP(i, n){
        cin >> b[i];
    }
    as = bs = 0;
    int flag = 0;
    REP(i, n){
        if(a[i] == b[i] || (a[i] < b[i] && (b[i] - a[i]) % 2 == 1)){
            flag += b[i] - a[i];
        }
        else if(a[i] > b[i]){
            bs += a[i] - b[i];
        }else{
            as += (b[i] - a[i]) / 2;
        }
    }
    if(flag >= as - bs && as - bs >= 0){
        cout << "Yes" << endl;
    }else{
        cout << "No" << endl;
    }
}