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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

//for
#define REP(i,n) for(int i=0,i##_len=(int)(n);i<i##_len;i++)
#define REPS(i,n) for(int i=1,i##_len=(int)(n);i<=i##_len;i++)
#define RREP(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define RREPS(i,n) for(int i=(int)(n);i>0;i--)

//printf
#define PRINTD(d) printf("%d\n",(d))
#define PRINTS(s) printf("%s\n",(s))

//memset
#define m0(s) memset(s,0,sizeof(s))
#define ml(s) memset(s,63,sizeof(s))
#define fill(s,c) memset(s,c,sizeof(s))

typedef long long ll;

int Min(int a, int b) {
	return (a)<(b) ? (a) : (b);
}
int Max(int a, int b) {
	return (a) > (b) ? (a) : (b);
}

int main() {
	int N;
	int a[10000], b[10000];
	int da=0, db=0;
	int dif = 0;
	scanf("%d", &N);
	REP(i, N) {
		scanf("%d", &a[i]);
	}
	REP(i, N) {
		scanf("%d", &b[i]);
	}
	REP(i, N) {
		if (a[i] > b[i]) {
			db += a[i] - b[i];
		}
		else if (a[i] < b[i]) {
			if ((b[i] - a[i]) % 2 == 0) {
				da += (b[i] - a[i]) / 2;
			}
			else {
				da += ((b[i] + 1) - a[i]) / 2;
				db++;
			}
		}
	}

	if (da>=db) {
		PRINTS("Yes");
	}
	else {
		PRINTS("No");
	}
}