#define _CRT_SECURE_NO_WARNINGS 1
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
#include<random>
#include <chrono>
using namespace std;
std::mt19937_64 rng((std::chrono::steady_clock::now().time_since_epoch().count()));
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define RND mt19937 rnd(0xab8d5f6);
#define time_while(time_while_val) while (((double)clock() - start) / clocks_per_sec < time_while_val)
//#define time_while(time_while_val) while ((double)clock() / clocks_per_sec < time_while_val)//弃
#define rep1(i, n) for(long long i = 1; i <= n; ++i)
#define rep0(i, n) for(long long i = 0; i <= n; ++i)
#define double long double
//#define int long long
//#define ll unsigned long long
#define lc u<<1
#define rc u<<1|1
#define endl "\n"
#define X first
#define Y second
typedef long long ll;
//typedef unsigned long long ull;
typedef pair<int, int> PII;
double PI = acos(-1.0);
const double eps = 1e-9;
const ll maxn = 2e6 + 5;
const ll maxm = 2e18;
const ll inf = 2147483647;
const ll mod = 1e9 + 7;
const int dx[] = { 1,0,-1,0,1,1,-1,-1 };//前4为邻
const int dy[] = { 0,-1,0,1,-1,1,-1,1 };

inline int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    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;
}
int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}
ll ksm(ll a, ll b) {
    ll ans = 1;
    ll base = a;
    while (b) {
        if (b & 1)ans = ans * base % mod;
        base = base * base % mod;
        b >>= 1;
    }
    return ans % mod;
}
ll getinv(ll a) {
    return ksm(a, mod - 2);
}
void debug(ll ans) {
    cout << "业师大人帮你检查答案：" << ans << endl;
}

/*
ll getinv(ll a) {
    return ksm(a, mod - 2);
}*/

//dismyth
//警钟长鸣

/*2024/8/6
待补:cdq分治补不出来，三维偏序，平衡树->link cut tree*/
/*待补 324(包括f),310，313,315,317*///好多，但是每天得vp，强度得上去
/*306d,344d*/
//赛时多出，赛后少补
//vp一把


/*
struct cmp {
    bool operator()(pair<ll, ll> a, pair<ll, ll>  b) {
        //return a.X > b.X;//表示小根堆

    }
};*/

//异或是可叠加的，所以异或是可求前缀和的
//注意map初始化为0遇到ans为0时记忆化会失效，这时候我们就要用到.count
//re.count({l,r})

int a[maxn];
int n, m;
bool vis[maxn];
int ans[maxn];
vector<int>mid;

void dfs(int now) {
    mid.push_back(a[now]);
    if (vis[now] || now > n) {//走过了，我们，依然要向后走
        int mm = now;
        int cnt = 0;
        while (mm < now + n) {
            int pos = lower_bound(a + 1, a + 1 + 2 * n, a[mm] + m) - a;
            if (pos == 1 + 2 * n) {
                cnt++;
                mid.push_back(1e9 + 5); break;
            }
            mid.push_back(a[pos]);
            mm = pos;
            cnt++;
        }
        ans[now] = cnt;
        if (a[now] - a[now - 1] >= m)ans[now]++;
        return;
    }//退出
    vis[now] = 1;
    int pos = lower_bound(a + 1, a + 1 + 2 * n, a[now] + m) - a;
    //mid.push_back(a[pos]);
    dfs(pos);
    ans[now] = (lower_bound(mid.begin(), mid.end(), a[now+n-1])) - (lower_bound(mid.begin(), mid.end(), a[now]));
    if (a[now] - a[now - 1] >= m) {
        ans[now]++;
    }
    return;
}


//RND;
void solve() {
    //我应该直接走，就是，对于每一个位置，我们都尝试去向前走

    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> a[i]; a[i + n] = a[i];
    }
    a[0] = 0;
    for (int i = 1; i <= 2 * n; i++) {
        a[i] += a[i - 1]; vis[i] = 0;
    }
    vis[0] = 0;
    //获取前缀和，然后对于其中的元素，没走过就一直走
    //但是我们怎么记录ans
    for (int i = 1; i <= n; i++) {
        if (vis[i] == 0) {
            dfs(i);
            mid.clear();
        }
    }
    for (int i = 1; i <= n; i++) {
        cout << ans[i] - 1 << " ";
    }
    cout << endl;
    return;
}


signed main() {
    IOS;
    ll t = 1;
    //cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

/*致我已死的友人*/
/*PonyHex*/