codeforces round#522 E - The Unbearable Lightness of Weights

You have a set of nn weights. You know that their masses are a1a1, a2a2, ..., anan grams, but you don't know which of them has which mass. You can't distinguish the weights.

However, your friend does know the mass of each weight. You can ask your friend to give you exactly kkweights with the total mass mm (both parameters kk and mm are chosen by you), and your friend will point to any valid subset of weights, if it is possible.

You are allowed to make this query only once. Find the maximum possible number of weights you can reveal after this query.

Input

The first line contains a single integer nn (1≤n≤1001≤n≤100) — the number of weights.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100) — the masses of the weights.

Output

Print the maximum number of weights you can learn the masses for after making a single query.

Examples

Input

4
1 4 2 2

Output

2

Input

6
1 2 4 4 4 9

Output

2

Note

In the first example we can ask for a subset of two weights with total mass being equal to 44, and the only option is to get {2,2}{2,2}.

Another way to obtain the same result is to ask for a subset of two weights with the total mass of 55 and get {1,4}{1,4}. It is easy to see that the two remaining weights have mass of 22 grams each.

In the second example we can ask for a subset of two weights with total mass being 88, and the only answer is {4,4}{4,4}. We can prove it is not possible to learn masses for three weights in one query, but we won't put the proof here.

题意:有n个砝码,你不知道每个砝码的重量,但是你的朋友知道,你可以每次拿出k个砝码给你的朋友,他会告诉你k个砝码的重量和,经过多次尝试之后,你最多可以区分出多少个砝码？

题解:首先对于你拿出的k个砝码,若要想区分出来,这k个砝码必然是相同的且其质量和的方案数唯一,那么对于所有砝码的重量和的方案数可以用多重背包来求解,最后特判只有2种物品的就可以了。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e4+7;

int a[maxn],dp[maxn][105];
int main()
{
    int n;
    scanf("%d",&n);
    map<int,int>m;

    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        m[ a[i] ]++;
    }

    vector<pair<int,int> >v;

    for(auto p:m) v.push_back(p);

    dp[0][0]=1;
    for(auto p:v)
    {
        int vol=p.first;
        int k=p.second;
        for(int i=1e4;i>0;i--)
            for(int j=1;j<=n;j++)
                for(int t=1;t<=k && t<=j && i-vol*t>=0;t++)
                {
                    dp[i][j]+=dp[i-vol*t][j-t];
                    dp[i][j]=min(dp[i][j],2);
                }
    }

    int ans=1;
    for(auto p:v)
    {
        int vol=p.first;
        int k=p.second;
        for(int i=1;i<=k;i++)
            if(dp[i*vol][i]==1)
            {
                if(v.size()==2 && i==k)
                {
                    printf("%d\n",n);
                    return 0;
                }
                else ans=max(ans,i);
            }
    }

    printf("%d\n",ans);
    return 0;
}