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: Time Limit Exceeded)

#include<stdio.h>

long min(long a,long b){
  if(a<b) return a;
  else return b;
}

int main(){
  int i,j,n,flag[2]={},ii=0,jj=0;
  long a[10000],b[10000],sa=0,sb=0,c;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%ld",&a[i]);
    sa+=a[i];
  }
  for(i=0;i<n;i++){
    scanf("%ld",&b[i]);
    sb+=b[i];
  }

  if(sa>sb) printf("No\n");
  else{
    do{
      flag[0]=flag[1]=0;
      for(i=ii;i<n;i++){
        if(a[i]<b[i]){
          flag[0]=1;
          break;
        }
      }
      for(j=jj;j<n;j++){
        if(a[j]>b[j]||(a[j]-b[j])%2==1){
          flag[1]=1;
          break;
        }
      }
      if(a[j]>b[j]) c=min((b[i]-a[i])/2,(a[j]-b[j]));
      else c=1;
      if(flag[0]){
        a[i]+=2*c;
        b[j]+=c;
        sa+=2*c;
        sb+=c;
      }
    ii=i;
    jj=j;
    }while(flag[0]&&sa<sb);
    flag[0]=1;
    for(i=0;i<n;i++){
      if(a[i]!=b[i]){
        flag[0]=0;
        break;
      }
    }
    if(flag[0]) printf("Yes\n");
    else printf("No\n");
  }
}