Problem: Misha has an array of integers of length n. He wants to choose k different continuous subarrays, so that each element of the array belongs to at least one of the chosen subarrays.

Misha wants to choose the subarrays in such a way that if he calculated the sum of elements for each subarray, and then add up all these sums, the resulting value was maximum possible.

Input

The first line of input contains two integers: n, k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ n·(n + 1) / 2) — the number of elements in the array and the number of different subarrays that must be chosen.

The second line contains n integers ai ( - 50 000 ≤ ai ≤ 50 000) — the elements of the array.

Output

Output one integer — the maximum possible value Misha can get by choosing k different subarrays.

Example

Input

5 4
6 -4 -10 -4 7


Output

11

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long k, qzh[100005], val[100005], l, r, mid, ans;
long long n, i, j, a[100005], seq[100005], rk[100005], rt[100005], lc[5000005],
    rc[5000005];
long long mx[100005][19], mi[100005][19];
long long lg[100005], cnt, pr[100005], cov[100005];
bool cmp(long long x, long long y) { return qzh[x] < qzh[y]; }
long long querymx(long long l, long long r) {
  long long ans = -0x3f3f3f3f3f3f3f3fll;
  if (l < 1) {
    ans = 0;
    l++;
  }
  if (l <= r) {
    long long t = lg[r - l + 1];
    return max(ans, max(mx[l][t], mx[r - (1 << t) + 1][t]));
  }
  return ans;
}
long long querymi(long long l, long long r) {
  long long ans = 0x3f3f3f3f3f3f3f3fll;
  if (l < 1) {
    ans = 0;
    l++;
  }
  if (l <= r) {
    long long t = lg[r - l + 1];
    return min(ans, min(mi[l][t], mi[r - (1 << t) + 1][t]));
  }
  return ans;
}
struct seg {
  long long sum;
  long long sz, mx;
} s[5000005];
void update(long long &x, long long l, long long r, long long y) {
  cnt++;
  s[cnt] = s[x];
  lc[cnt] = lc[x];
  rc[cnt] = rc[x];
  x = cnt;
  s[x].sz++;
  s[x].mx = max(s[x].mx, seq[y]);
  s[x].sum += val[y];
  if (l == r) return;
  long long mid = (l + r) / 2;
  if (y <= mid)
    update(lc[x], l, mid, y);
  else
    update(rc[x], mid + 1, r, y);
}
seg tmp;
void query(long long x, long long l, long long r, long long v) {
  if (!x) return;
  if (val[l] > v) {
    tmp.mx = max(tmp.mx, s[x].mx);
    tmp.sz += s[x].sz;
    tmp.sum += s[x].sum;
    return;
  }
  if (l == r) return;
  long long mid = (l + r) / 2;
  if (v < val[mid]) query(lc[x], l, mid, v);
  query(rc[x], mid + 1, r, v);
}
struct ii {
  long long ans;
  long long cnt;
} dp[100005];
struct itv {
  long long l, r;
} ss[100005];
ii solve(long long d) {
  ii ans = (ii){0, 0};
  long long i;
  memset(pr, 0, sizeof(pr));
  for ((i) = (n); (i) >= 1; --(i)) {
    tmp = (seg){0, 0, 0};
    query(rt[i], 1, n, qzh[i - 1] - d);
    pr[i] = tmp.mx;
    ans.ans += tmp.sum - (qzh[i - 1] - d) * tmp.sz;
    ans.cnt += tmp.sz;
  }
  memset(cov, 0, sizeof(cov));
  long long lst = 0;
  for ((i) = 1; (i) <= (n); ++(i)) {
    if (pr[i] > lst) {
      lst = max(lst, i - 1);
      while (lst < pr[i]) {
        cov[++lst] = 1;
      }
    }
  }
  long long cc = 0;
  if (!cov[1]) ss[++cc] = (itv){1, 0};
  for ((i) = 1; (i) <= (n); ++(i)) {
    if (cov[i]) {
      if (i == 1 || !cov[i - 1]) ss[++cc] = (itv){i, 0};
      if (i == n || !cov[i + 1]) ss[cc].r = i;
    } else {
      if (i > 1 && !cov[i - 1]) {
        ss[++cc] = (itv){i + 1, i};
      }
    }
  }
  if (!cov[n]) ss[++cc] = (itv){n + 1, n};
  if (cc == 1) return ans;
  ii mx = (ii){-0x3f3f3f3f3f3f3f3fll, 0};
  for ((i) = 1; (i) <= (cc); ++(i)) {
    if (i == 1)
      dp[i] = (ii){0, 0};
    else {
      dp[i].ans = mx.ans + querymx(ss[i].l - 1, ss[i].r) + d;
      dp[i].cnt = mx.cnt + 1;
    }
    ii t;
    t.ans = dp[i].ans - querymi(ss[i].l - 1, ss[i].r);
    t.cnt = dp[i].cnt;
    if (t.ans > mx.ans || t.ans == mx.ans && t.cnt < mx.cnt) {
      mx = t;
    }
  }
  return (ii){ans.ans + dp[cc].ans, ans.cnt + dp[cc].cnt};
}
signed main() {
  for ((i) = (2); (i) <= (100000); ++(i)) lg[i] = lg[i / 2] + 1;
  scanf("%lld%lld", &n, &k);
  for ((i) = 1; (i) <= (n); ++(i)) {
    scanf("%lld", &a[i]);
    qzh[i] = qzh[i - 1] + a[i];
    seq[i] = i;
  }
  sort(seq + 1, seq + n + 1, cmp);
  for ((i) = 1; (i) <= (n); ++(i)) {
    rk[seq[i]] = i;
    val[i] = qzh[seq[i]];
  }
  for ((i) = (n); (i) >= 1; --(i)) {
    rt[i] = rt[i + 1];
    update(rt[i], 1, n, rk[i]);
  }
  for ((i) = 1; (i) <= (n); ++(i)) mi[i][0] = mx[i][0] = qzh[i];
  for ((j) = 1; (j) <= (18); ++(j))
    for ((i) = 1; (i) <= (n - (1 << j) + 1); ++(i)) {
      mi[i][j] = min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]);
      mx[i][j] = max(mx[i][j - 1], mx[i + (1 << (j - 1))][j - 1]);
    }
  solve(3);
  l = -1e15;
  r = 1e15;
  while (l < r) {
    mid = (l + r) >> 1;
    ii s = solve(mid);
    if (s.cnt <= k) {
      l = mid + 1;
      ans = s.ans - 1ll * k * mid;
    } else {
      r = mid;
    }
  }
  printf("%lld\n", ans);
  return 0;
}