You have a set of nn weights. You know that their masses are a1, a2, ..., 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 k weights with the total mass m (both parameters k 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 n (1≤n≤1001≤n≤100) — the number of weights.

The second line contains nn integers na1,a2,…,an (1≤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.

Input

4
1 4 2 2
2
Input

6
1 2 4 4 4 9
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}.

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}. It is easy to see that the two remaining weights have mass of 2 grams each.

In the second example we can ask for a subset of two weights with total mass being 8, and the only answer is {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.

題意:

給你一個n表示物品的數量，然後跟著n個物品的重量，有一個k表示每次可以知道k個物品的重量的和，問你最大可以知道幾個物品確切的重量

複習了一波多重揹包，注意特判2種 的情況

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e4+7;
int x,dp[maxn][105],n;
vector<pair<int,int> > vec;
map<int,int> vis;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&x);
        vis[x]++;
    }
    for(auto p:vis) vec.push_back(p);/// pair <x,x.cnt>
    if(vec.size()==2) ///特判只有兩種質量物品
        return 0*printf("%d\n",n);
    dp[0][0]=1;
    for(auto p:vec){
        int vol=p.first;///物體的權-->質量
        int k=p.second;///個數
        for(int i=1e4;i;i--)///揹包容積--->質量
            for(int j=1;j<=n;j++)///放j件時
                for(int t=1; (t<=k && t<=j && i-vol*t>=0) ;t++)///更新方案數，放j件是從第i-1個物體轉移過來
                    dp[i][j] += dp[i-vol*t][j-t];
    }
    int ans=1;
    for(auto p:vec){
        int vol = p.first;
        int k = p.second;
        for(int i=1;i<=k;i++)
            if(dp[i*vol][i]==1)///方案唯一則可以判斷
                ans=max(ans,i);
    }
    printf("%d\n",ans);
    return 0;
}