Problem: The prestigious Codeforces kindergarten consists of n kids, numbered 1 through n. Each of them are given allowance in rubles by their parents.

Today, they are going to the most famous candy shop in the town. The shop sells candies in packages: for all i between 1 and m, inclusive, it sells a package containing exactly i candies. A candy costs one ruble, so a package containing x candies costs x rubles.

The kids will purchase candies in turns, starting from kid 1. In a single turn, kid i will purchase one candy package. Due to the highly competitive nature of Codeforces kindergarten, during a turn, the number of candies contained in the package purchased by the kid will always be strictly greater than the number of candies contained in the package purchased by the kid in the preceding turn (an exception is in the first turn: the first kid may purchase any package). Then, the turn proceeds to kid i + 1, or to kid 1 if it was kid n's turn. This process can be ended at any time, but at the end of the purchase process, all the kids must have the same number of candy packages. Of course, the amount spent by each kid on the candies cannot exceed their allowance.

You work at the candy shop and would like to prepare the candies for the kids. Print the maximum number of candies that can be sold by the candy shop to the kids. If the kids cannot purchase any candy (due to insufficient allowance), print 0.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 2·105, 2 ≤ m ≤ 5·106, n ≤ m), denoting the number of kids and the maximum number of candies in a package sold by the candy shop, respectively.

Then n lines follow, each line will contain a single positive integer not exceeding <image> denoting the allowance of a kid in rubles. The allowances are given in order from kid 1 to kid n.

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is recommended to use cin, cout streams (also you may use %I64d specificator).

Output

Print a single integer denoting the maximum number of candies that can be sold by the candy shop.

Examples

Input

2 5
5
10


Output

13


Input

3 8
8
16
13


Output

32


Input

2 5000000
12500002500000
12500002500000


Output

12500002500000

Note

For the first example, one of the scenarios that will result in 13 purchased candies is as follows. 

  * Turn 1. Kid 1 purchases 1 candy. 
  * Turn 2. Kid 2 purchases 3 candies. 
  * Turn 3. Kid 1 purchases 4 candies. 
  * Turn 4. Kid 2 purchases 5 candies. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000 + 1000;
const long long INF = 0x1f1f1f1f1f1f1f1fll;
struct Node {
  int l, r;
  pair<long long, long long> s;
  long long p;
  Node *lc, *rc;
  Node(int _l = 0, int _r = 0)
      : l(_l), r(_r), s(INF, 0), p(0), lc(NULL), rc(NULL) {}
} * root;
int n, m, r = 0;
long long g[MAXN], a[MAXN], ans = 0;
pair<long long, long long> merge(const pair<long long, long long> &a,
                                 const pair<long long, long long> &b) {
  if (a.first == b.first)
    return make_pair(a.first, max(a.second, b.second));
  else if (a.first < b.first)
    return a;
  else
    return b;
}
void add(Node *now, long long v) {
  now->p += v;
  now->s.first += v;
}
void down(Node *now) {
  if (now->p) {
    add(now->lc, now->p);
    add(now->rc, now->p);
    now->p = 0;
  }
}
void update(Node *now) { now->s = merge(now->lc->s, now->rc->s); }
void build(Node *&now, int l, int r) {
  now = new Node(l, r);
  if (l == r) return;
  int mid = (l + r) / 2;
  build(now->lc, l, mid);
  build(now->rc, mid + 1, r);
}
void rebuild(Node *now, int l, int r) {
  now->s = make_pair(INF, 0);
  now->p = 0;
  if (l == r) {
    now->s = make_pair(a[l], l);
    return;
  }
  int mid = (l + r) / 2;
  rebuild(now->lc, l, mid);
  rebuild(now->rc, mid + 1, r);
  update(now);
}
void modify(Node *now, int l, int r, long long v) {
  if (now->r < l || r < now->l) return;
  if (l <= now->l && now->r <= r) {
    add(now, v);
    return;
  }
  down(now);
  modify(now->lc, l, r, v);
  modify(now->rc, l, r, v);
  update(now);
}
pair<long long, long long> query(Node *now, int l, int r) {
  if (now->r < l || r < now->l) return make_pair(INF, 0);
  if (l <= now->l && now->r <= r) return now->s;
  down(now);
  return merge(query(now->lc, l, r), query(now->rc, l, r));
}
long long calc(int r) {
  long long ret = 0;
  long long lef = m - n * r;
  for (int i = 1; i <= n; i++) {
    a[i] = g[i] - ((long long)(r - 1) * r * n / 2 + r * i);
    ret += ((long long)(r - 1) * r * n / 2 + r * i);
    if (a[i] < 0) return -1;
  }
  rebuild(root, 1, n);
  while (lef >= 1) {
    pair<long long, long long> res = query(root, 1, n);
    long long p = res.second, v = res.first;
    if (p == n || v >= lef * r) {
      if (v == 0)
        break;
      else if (v >= r) {
        long long times = min(lef, v / r);
        modify(root, 1, n, -(r * times)), ret += r * n * times;
        lef -= times;
      } else {
        modify(root, 1, n, -v), ret += v * n;
        lef -= 1;
      }
    } else {
      if (v == 0) {
        pair<long long, long long> res = query(root, p + 1, n);
        long long p2 = res.second, v2 = res.first;
        long long times = min(lef, v2);
        modify(root, p + 1, n, -(1 * times)), ret += (n - p) * 1 * times;
        lef -= times;
      } else if (v >= r - 1) {
        pair<long long, long long> res = query(root, p + 1, n);
        long long p2 = res.second, v2 = res.first;
        long long times = lef;
        if (r - 1 > 0) times = min(times, v / (r - 1));
        min(times, v2 / r);
        times = min(times, v2 - v);
        times = min(times, lef * r - v);
        modify(root, 1, p, -((r - 1) * times)), ret += p * (r - 1) * times;
        modify(root, p + 1, n, -(r * times)), ret += (n - p) * r * times;
        lef -= times;
      } else {
        modify(root, 1, p, -v), ret += p * v;
        modify(root, p + 1, n, -(v + 1)), ret += (n - p) * (v + 1);
        lef -= 1;
      }
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> g[i];
  build(root, 1, n);
  for (int i = 1; i * n <= m; i++) {
    long long res = calc(i);
    if (res < 0) break;
    ans = max(ans, res);
  }
  cout << ans << endl;
  return 0;
}