#include <bits/stdc++.h>
using namespace std;
typedef priority_queue<pair<int, pair<int, int> >,
                       vector<pair<int, pair<int, int> > >,
                       greater<pair<int, pair<int, int> > > >
    pqset_ll;
long long int dxk[] = {0, 0, 1, -1, 1, 1, -1, -1};
long long int dyk[] = {1, -1, 0, 0, 1, -1, 1, -1};
long long int ll_max(long long int a, long long int b, long long int c) {
  return max(a, max(b, c));
}
int int_max(int a, int b, int c) { return max(a, max(b, c)); }
long long int ll_min(long long int a, long long int b, long long int c) {
  return min(a, min(b, c));
}
int int_min(int a, int b, int c) { return min(a, min(b, c)); }
long long int max(int a, long long int b) { return max((long long int)a, b); }
long long int min(int a, long long int b) { return min((long long int)a, b); }
long long int min(long long int a, int b) { return min(a, (long long int)b); }
long long int max(long long int a, int b) { return max(a, (long long int)b); }
long long int dx[] = {0, 0, 1, -1};
long long int dy[] = {1, -1, 0, 0};
long long int power(long long int a, long long int b) {
  if (a == 1) return 1;
  if (b == 0) return 1;
  long long int c = power(a, b / 2);
  long long int res = 1;
  if (b % 2) {
    res = (c * c) % 1000000007;
    res *= a;
    res %= 1000000007;
  } else
    res = ((c * c) % 1000000007);
  return res;
}
long long int modInv(long long int a) {
  return power(a, 1000000007 - 2) % 1000000007;
}
long long int fact[1], inv[1];
void factorial(long long int n) {
  fact[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= 1000000007;
  }
}
void InvFactorial(long long int n) {
  inv[0] = 1;
  for (long long int i = 1; i <= n; i++) inv[i] = modInv(fact[i]);
}
long long int ncr(long long int n, long long int r) {
  if (n < r || n < 0 || r < 0) return 0;
  long long int b = inv[n - r];
  long long int c = inv[r];
  long long int a = fact[n] * b;
  a %= 1000000007;
  a *= c;
  a %= 1000000007;
  return a;
}
long long int Tree[1];
void build_max(int node, int start, int end, int a[]) {
  if (start == end) {
    Tree[node] = a[start];
    return;
  }
  int mid = ((start + end) >> 1);
  build_max(2 * node + 1, start, mid, a);
  build_max(2 * node + 2, mid + 1, end, a);
  Tree[node] = max(Tree[2 * node + 1], Tree[2 * node + 2]);
}
void update_max(int node, int start, int end, int i, long long int val) {
  if (start == end) {
    Tree[node] = val;
    return;
  }
  int mid = ((start + end) >> 1);
  if (i <= mid)
    update_max(2 * node + 1, start, mid, i, val);
  else
    update_max(2 * node + 2, mid + 1, end, i, val);
  Tree[node] = max(Tree[2 * node + 1], Tree[2 * node + 2]);
}
long long int query_max(int node, int start, int end, int l, int r) {
  if (start > end || end < l || start > r) return -100000000;
  if (start >= l && end <= r) return (Tree[node]);
  int mid = (start + end) / 2;
  long long int p1 = query_max(node * 2 + 1, start, mid, l, r);
  long long int p2 = query_max(node * 2 + 2, mid + 1, end, l, r);
  return max(p1, p2);
}
long long int s(long long int l, long long int r) {
  return (r * (r + 1)) / 2 - ((l - 1) * l) / 2;
}
void solve(int countu) {
  int n;
  long long int x;
  cin >> n >> x;
  long long int a[n];
  vector<long long int> pref(2 * n, 0), sum(2 * n, 0);
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (i == 0) {
      pref[i] = a[i];
      sum[i] = (a[i] * (a[i] + 1)) / 2;
    } else {
      pref[i] = pref[i - 1] + a[i];
      sum[i] = (a[i] * (a[i] + 1)) / 2 + sum[i - 1];
    }
    if (a[i] >= x) ans = max(ans, s(a[i] - x + 1, a[i]));
  }
  for (long long int i = n; i < 2 * n; i++) {
    pref[i] = pref[i - 1] + a[i - n];
    sum[i] = sum[i - 1] + s(1, a[i - n]);
  }
  for (long long int i = 0; i < n; i++) {
    long long int p = 0;
    if (a[i] > x) continue;
    if (i > 0) p = pref[i - 1];
    long long int j =
        lower_bound(pref.begin(), pref.end(), p + x) - pref.begin();
    if (j == i) continue;
    long long int k = p + x - pref[j - 1];
    long long int t =
        sum[j - 1] - ((i - 1 < 0) ? 0 : sum[i - 1]) + (k * (k + 1)) / 2;
    long long int o = 0;
    k = pref[j - 1] - pref[i];
    k = x - k;
    if (k > pref[j] - pref[j - 1]) {
      k -= pref[j] - pref[j - 1];
      o = sum[j - 1] - sum[i] + s(a[i] - k + 1, a[i]) +
          s(1, pref[j] - pref[j - 1]);
    }
    ans = ll_max(ans, t, o);
  }
  cout << ans << '\n';
}
int main() {
  int t = 1;
  int countu = 1;
  while (t--) {
    solve(countu);
    countu++;
  }
}
