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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] a = new int[n];
		int[] b = new int[n];
		for(int i=0; i<n; i++) {
			a[i] = sc.nextInt();
		}
		for(int i=0; i<n; i++) {
			b[i] = sc.nextInt();
		}
		Arrays.sort(a); Arrays.sort(b);
		for(int i=0; i<n; i++) {
			if(a[i]<=b[i]) {
				
			} else {
				System.out.println("No"); return;
			}
		}
		System.out.println("Yes");
	}
	
	static int binarySearch(int[] A, int x) {
		int l = 0;
		int r = A.length;
		while(l<r) {
			int mid = (l+r)/2;
			if(A[mid] == x) return mid;
			else if(x<A[mid]) r = mid;
			else l = mid+1;
		}
		return -1;
	}
	
	static long upper_bound(int[] A, int b) {
		// bよりおおきい値がはじめて出てくるindexを返す
		return lower_bound(A, b+1);
	}

	static long lower_bound(int[] A, int b) {
		// b以上の値がはじめて出てくるindexを返す
		int lb=0;
		int ub=A.length;
		while(ub>lb) {
			int mid = (lb+ub)/2;
			if(A[mid]>=b) ub = mid;
			else lb = mid+1;
		}
		return lb;
	}
	
	// union find lib
	// usage:
	// 最初にinitを呼ぶ
	// root: 直接は呼ばないで
	// unite: まとめる
	// same: グループ判定
	static void init(int par[], int N) {
		for(int i=0; i<N; i++) {
			par[i] = i;
		}
	}
	
	static int root(int x, int [] par) {
		if(par[x] == x) {
			return x;
		} else {
			return (par[x] = root(par[x], par));
		}
	}
	
	static boolean same(int x, int y, int[] par) {
		return root(x, par) == root(y, par);
	}
	
	static void unite(int x, int y, int[] par) {
		x = root(x, par);
		y = root(y, par);
		if(x == y) return;
		par[x] = y;
	}
	
	// end union find lib
	
	// number lib
	
	static int max(int...ls) {
		int ans = Integer.MIN_VALUE;
		for(int i=0; i<ls.length; i++) {
			ans = Math.max(ans, ls[i]);
		}
		return ans;
	}
	
	static int min(int...ls) {
		int ans = Integer.MAX_VALUE;
		for(int i=0; i<ls.length; i++) {
			ans = Math.min(ans, ls[i]);
		}
		return ans;
	}
	
	static long lcm(long a, long b) {
		return a*(b/gcd(a, b));
	}
	
	static long gcd(long a, long b) {
		long ta = Math.max(a, b);
		long tb = Math.min(a, b);
		long tmp;
		while((tmp = ta%tb) != 0) {
			ta = tb;
			tb = tmp;
		}
		return tb;
	}
	
	static long modcomb(long n, long k, long mod) {
		if(k==1) {
			return n;
		}
		
		long ans = 1;
		for(long i=n; i>=n-k+1; i--) {
			ans = (ans * i)%mod;
		}
		for(long i=k; 0<i; i--) {
			ans = (ans * modpow(i, mod-2, mod)) % mod;
		}
		return ans;
	}
	
	static long modpow(long a, long b, long mod) {
		if(b==0) return 1;
		if(b%2==0) {
			long d = modpow(a, b/2, mod);
			return (d*d)%mod;
		} else {
			return (a*modpow(a, b-1, mod))%mod;
		}
	}
	
	static int disit(long a, long d) {
		int count = 0;
		while(a!=0) {
			a = a/d;
			count++;
		}
		return count;
	}
	
	// end number lib
}