You are given array a_1, a_2, ..., a_n. You need to split it into k subsegments (so every element is included in exactly one subsegment).

The weight of a subsegment a_l, a_{l+1}, ..., a_r is equal to (r - l + 1) ⋅ max_{l ≤ i ≤ r}(a_i). The weight of a partition is a total weight of all its segments.

Find the partition of minimal weight.

Input

The first line contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^4, 1 ≤ k ≤ min(100, n)) — the length of the array a and the number of subsegments in the partition.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2 ⋅ 10^4) — the array a.

Output

Print single integer — the minimal weight among all possible partitions.

Examples

Input


4 2
6 1 7 4


Output


25


Input


4 3
6 1 7 4


Output


21


Input


5 4
5 1 5 1 5


Output


21

Note

The optimal partition in the first example is next: 6 1 7 \bigg| 4.

The optimal partition in the second example is next: 6 \bigg| 1 \bigg| 7 4.

One of the optimal partitions in the third example is next: 5 \bigg| 1 5 \bigg| 1 \bigg| 5.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pt = pair<ll, ll>;
const int MXN = 2e4;
const int MXM = 1e2;
const int INF = 1e9;
pt operator-(pt a, pt b) { return {a.first - b.first, a.second - b.second}; }
pt operator*(pt a, pt b) {
  return {a.first * b.first - a.second * b.second,
          a.first * b.second + a.second * b.first};
}
ll dot(pt a, pt b) { return a.first * b.first + a.second * b.second; }
ll cross(pt a, pt b) { return a.first * b.second - a.second * b.first; }
struct cht {
  deque<pt> hull, vecs;
  void addfront(pt ln) {
    while (!vecs.empty() && dot(vecs.front(), ln - hull.front()) < 0) {
      hull.pop_front();
      vecs.pop_front();
    }
    if (!hull.empty()) {
      vecs.push_front((hull.front() - ln) * pt{0, 1});
    }
    hull.push_front(ln);
  }
  void addback(pt ln) {
    while (!vecs.empty() && dot(vecs.back(), ln - hull.back()) < 0) {
      hull.pop_back();
      vecs.pop_back();
    }
    if (!hull.empty()) {
      vecs.push_back((ln - hull.back()) * pt{0, 1});
    }
    hull.push_back(ln);
  }
  ll qry(int first) {
    auto it = lower_bound(vecs.begin(), vecs.end(), pt{first, 1},
                          [](pt a, pt b) { return cross(a, b) > 0; });
    return dot(hull[it - vecs.begin()], {first, 1});
  }
};
int chd[MXN * 100][2], a[MXN + 1], sz, n;
ll dp[2][MXN + 1];
pt lct[MXN * 100];
int newnode(int chd0, int chd1, pt ln) {
  int u = ++sz;
  chd[u][0] = chd0;
  chd[u][1] = chd1;
  lct[u] = ln;
  return u;
}
int upd(pt ln, int u, int l = 1, int r = n) {
  if (!u) return newnode(0, 0, ln);
  int m = (l + r) / 2;
  bool mnl = dot(ln, {l, 1}) < dot(lct[u], {l, 1});
  bool mnm = dot(ln, {m, 1}) < dot(lct[u], {m, 1});
  pt res = lct[u];
  if (mnm) {
    swap(res, ln);
  }
  if (l == r) return newnode(0, 0, res);
  if (mnl == mnm) return newnode(chd[u][0], upd(ln, chd[u][1], m + 1, r), res);
  return newnode(upd(ln, chd[u][0], l, m), chd[u][1], res);
}
ll qry(int first, int u, int l = 1, int r = n) {
  if (!u) return INF;
  int m = (l + r) / 2;
  ll res = dot(lct[u], {first, 1});
  if (l == r) return res;
  if (first <= m) return min(res, qry(first, chd[u][0], l, m));
  return min(res, qry(first, chd[u][1], m + 1, r));
}
void merge(cht& a, cht& b) {
  if (b.hull.size() > a.hull.size()) {
    for (auto it = a.hull.rbegin(); it != a.hull.rend(); ++it) {
      b.addfront(*it);
    }
    swap(a, b);
  } else {
    for (auto i : b.hull) {
      a.addback(i);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  fill(dp[0] + 1, dp[0] + 1 + n, INF);
  for (int i = 1; i <= k; ++i) {
    sz = 0;
    int r = i % 2;
    vector<cht> hulls;
    vector<int> mx, rt = {0};
    for (int j = 0; j <= n; ++j) {
      dp[r][j] = qry(j, rt.back());
      if (j == n) continue;
      cht cur;
      cur.addback({-j, dp[!r][j]});
      while (!mx.empty() && a[j] > mx.back()) {
        mx.pop_back();
        rt.pop_back();
        merge(cur, hulls.back());
        hulls.pop_back();
      }
      mx.push_back(a[j]);
      rt.push_back(upd({a[j], cur.qry(a[j])}, rt.back()));
      hulls.push_back(move(cur));
    }
  }
  cout << dp[k % 2][n] << '\n';
}
