/*
标称并没有用到题解中提到的那种奇妙的性质，我们可以证明，正常从1开始走的话，需要T次，如何使这个次数减小？题解中提到一个办法，有一步小于n/t，我们考虑这一步，如果把它匀到左右两步中，则可以减小，就根据这个性质来优化
next函数的部分，我当时用了一个倍增法，题解用了一个并查集，倍增比较直观，然而并查集更为巧妙
*/
//my code
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define ll long long
using namespace std;
const int maxn = 600500;
int n,q,cnt,ct[maxn];
ll a[maxn],sum[maxn],ans;
ll read(){
    char ch=getchar();
    ll x=0,f=1;
    while(!(ch>=‘0‘&&ch<=‘9‘)){if(ch==‘-‘)f=-1;ch=getchar();};
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+(ch-‘0‘);ch=getchar();};
    return x*f;
}
ll greedy(int u,ll t){
    ll tot = 0,tmp = 0;
    ll j,lg,tst,nt;
    for(int i = 1;i <= n;i++){
        tmp++;
        if(a[u] + a[u+1] > t || i == n){
            
        }else{
            lg = 0;
            j = 1;
            tot = 0;
            while(lg>=0){
                if(u+j<=n){
                    tst = sum[u+j]-sum[u-1];
                    nt = u + j;
                }
                else{
                    tst = sum[n] - sum[u-1] + sum[u+j-n];
                    nt = u + j - n;    
                }
                if(tot + tst <= t){
                    i += j;
                    tot += tst - a[nt];
                    u = nt;
                    lg++;
                    j <<= 1;
                }else{
                    lg--;
                    j >>= 1;
                }
            }
        }
        u++;
        if(u > n) u = 1;
        if(tmp >= ans) break;
    }
    return tmp;
}
void work(ll t){
    cnt = 0;
    ans = maxn;
    ct[++cnt] = 1;
    ll sum = a[1];
    for(int i = n;i > 1;i--){
        sum += a[i];
        if(sum > t) break;
        ct[++cnt] = i;
    }
    for(int i = 1;i <= cnt;i++){
        ans = min(ans,greedy(ct[i],t));
    }
    cout<<ans<<endl;
}
int main(){
    freopen("stall.in","r",stdin);
    freopen("stall.out","w",stdout);
    n = read();
    q = read();
    for(int i = 1;i <= n;i++){
        a[i] = read();
        sum[i] = sum[i-1] + a[i];
    }
    ll b;
    for(int i = 1;i <= q;i++){
        b = read();
        work(b);
    }
    return 0;
}
//std
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cstdlib>
#include<string>
#include<bitset>
#define INF 1000000000
#define N 2000005
#define fi first
#define se second
#define debug(x) cout<<#x<<"="<<x<<endl
#define MP(x,y) make_pair(x,y)
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
LL a[N],s[N];
int to[N],fa[N],d[N];
int findr(int x)
{
    if(fa[x]==x)
        return x;
    else
    {
        int t=fa[x];
        fa[x]=findr(fa[x]);
        d[x]=d[x]+d[t];
        return fa[x];
    }
}

void Union(int x,int y)
{
    fa[x]=y;
    d[x]=1;
}

int main()
{
    int size = 256 << 20; // 256MB
    char *p = (char*)malloc(size) + size;
    __asm__("movl %0, %%esp\n" :: "r"(p));
    freopen("stall.in","r",stdin);
    freopen("stall.out","w",stdout);
    cin>>n>>q;
    for(i=1;i<=n;i++)
        scanf("%I64d",&a[i]),a[i+n]=a[i];
    for(i=n*2;i;i--)
        s[i]=s[i+1]+a[i];
    while(q--)
    {
        ans=INF;
        cin>>m;
        memset(d,0,sizeof(d));
        for(i=1;i<=n;i++)
            fa[i]=i,fa[i+n]=i+n;
        for(j=n*2,i=n*2;i;i--)
        {
            while(s[i]-s[j+1]>m) j--;
            to[i]=j+1;
            //debug(to[i]);
        }
        //if(q==1) return 0;
        for(i=1;i<=n;i++)
        {
            j=findr(i);
            while(to[j]<i+n)
            {
                k=to[j];
                Union(j,k);
                j=findr(k);
            }
            findr(i);
            ans=min(ans,d[i]);
        }
        cout<<ans+1<<endl;
    }
    return 0;
}
// davidlee1999WTK 2015/
// srO myk Orz
//ios::sync_with_stdio(false);