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<bits/stdc++.h>
using namespace std;
#define int long long
//#undef int
#define mk make_pair
#define pb push_back
typedef pair<int, int> pii;
const int mod = 1000000007;
const int INF = 1000000009;
const long long INFL = 1000000000000000018ll;

int N, a[10100], b[10100], sum1, sum2;

signed main()
{
    cin >> N;
    for(int i = 0; i < N; i++){
        cin >> a[i];
        sum1 += a[i];
    }
    for(int i = 0; i < N; i++){
        cin >> b[i];
        sum2 += b[i];
    }
    if(sum2 < sum1){
        printf("No\n");
        return 0;
    }
    int sa = sum2 - sum1;
    int x = sa;
    for(int i = 0; i < N; i++){
        if(a[i] < b[i]){
            x -= (b[i] - a[i] + 1) / 2;
            a[i] += ((b[i] - a[i] + 1) / 2) * 2;
        }
    }
    if(x < 0){
        printf("No\n");
        return 0;
    }
    int y = sa;
    for(int i = 0; i < N; i++){
        y -= a[i] - b[i];
    }
    if(x * 2 == y){
        printf("Yes\n");
    }
    else {
        printf("No\n");
    }




    return 0;
}
/*



*/