/*
一道一道有毒的题，忘了一种竟然可以n个线段
*/
#include<bits/stdc++.h>
using namespace std;
int d[200100][20];
int n,k,cur,ob,maxn;
void init()
{
    for (int j = 1; (1 << j) < n; j++){
        int t = (1 << j) - 1;
        for (int i = 0; i+t < n; i++){
            d[i][j] = max(d[i][j-1], d[i+(1<<(j-1))][j-1]);
        }
    }
}
inline int RMQ(int a, int b)
{
    int l = int(log(double(b-a+1))/log(2.0));
    return max(d[a][l], d[b+1-(1<<l)][l]);
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    while(scanf("%d%d",&n,&k)&&n>=0&&k>=0)
    {
        ob=0;
        maxn=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&d[i][0]);
            ob+=d[i][0];
            maxn=max(maxn,d[i][0]);
        }
        if(ob<=k)
        {
            puts("-1");
            continue;
        }
        init();//RMQ预处理
        /*二分找最小值*/
        int ans=n;
        for(int m=max(1,k/maxn);m<n;m++)
        {
            cur=0;
            int t=n/m;//分组的长度
            //cout<<"t="<<t<<endl;
            for(int i=1;i<=m;i++)//枚举的组数
            {
                cur+=RMQ(t*(i-1),t*i-1);
                //cout<<t*(i-1)+1<<" "<<t*i<<endl;
                if(cur>k)
                    break;
            }
            if(cur>k)
            {
                ans=m;
                break;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}