The conversion is 01 backpack CodeForces 433A-Kitahara Haruki & amp; #39; s Gift

Kitahara Haruki has boughtNApples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.

Each apple weights 100 grams or 200 grams. of course Kitahara Haruki doesn't want to offend any of his friend. therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.

But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. please, tell him: is it possible to divide all the apples in a fair way between his friends?

Input

The first line contains an integerN(1? ≤?N? ≤? 100)-the number of apples. The second line containsNIntegersW1 ,?W2 ,?...,?WN(WI? =? 100 orWI? =? 200), whereWIIs the weight ofI-Th apple.

Output

In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes ).

Sample test (s) input

3100 200 100

Output

YES

Input

4100 100 100 200

Output

NO

Note

In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa.

The question is that a person buys a lot of apple and then gives the quality of each apple. It is required to divide these apple into two sets of equal quality. Isn't it dynamic planning ???!!! Let's talk about the idea: First add all the quality to sum. If sum 2 is not zero, then NO is directly used, and sum/= 2; then there is a 01 backpack, as long as dp [sum/2] = sum/2; then, paste the Code:

#include
 
  #include
  
   #include
   
    #include
    
     using namespace std;int main(){    int i,j,k;    int t,n,m;    __int64 sum;    int liu[106],dp[250000];    while(scanf("%d",&n)!=EOF)    {        sum=0;        memset(liu,0,sizeof(liu));        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)        {            scanf("%d",&liu[i]);            sum+=liu[i];        }        if(sum%2!=0)        {            printf("NO\n");            continue;        }        sum/=2;        //printf("%d\n",sum);        for(i=1;i<=n;i++)        {            for(j=sum;j>=liu[i];j--)                dp[j]=max(dp[j],dp[j-liu[i]]+liu[i]);        }        if(dp[sum]==sum)            printf("YES\n");        else            printf("NO\n");    }    return 0;}
    
   
  
 

Let's talk about bugs. Thank you;