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;
template <class T>
bool minimize(T &x, const T &y) {
  if (x > y)
    x = y;
  else
    return false;
  return true;
}
template <class T>
bool maximize(T &x, const T &y) {
  if (x < y)
    x = y;
  else
    return false;
  return true;
}
struct Line {
  int x, y;
  Line(int x = 0, int y = 0) : x(x), y(y) {}
  friend bool bad(Line l1, Line l2, Line l3) {
    return 1ll * (l3.y - l1.y) * (l1.x - l2.x) <=
           1ll * (l2.y - l1.y) * (l1.x - l3.x);
  }
  friend int S(Line l, int val) { return l.x * val + l.y; }
};
const int N = 2e4 + 7, M = 107, oo = 2e9;
int n, m, t0 = 0, t1 = 1;
int a[N], f[2][N], suf[N], pre[N];
struct Convex_Hull {
  int cnt;
  Line Q[N];
  void add(Line l) {
    while (cnt && Q[cnt].x <= l.x) minimize(l.y, Q[cnt--].y);
    while (cnt >= 2 && bad(Q[cnt - 1], Q[cnt], l)) cnt--;
    Q[++cnt] = l;
  }
  int query(int x) {
    if (!cnt) return oo;
    int l = 1, r = cnt;
    while (l < r) {
      int mid = l + r >> 1;
      if (S(Q[mid], x) > S(Q[mid + 1], x))
        l = mid + 1;
      else
        r = mid;
    }
    return S(Q[l], x);
  }
  void clear() { cnt = 0; }
} CH;
void CalcSuf(int l, int r) {
  suf[r + 1] = 0;
  for (int i = r; i >= l; i--) suf[i] = max(a[i], suf[i + 1]);
}
void CalcPre(int l, int r) {
  pre[l - 1] = 0;
  for (int i = l; i <= r; i++) pre[i] = max(a[i], pre[i - 1]);
}
void DaC(int l, int r) {
  int mid = l + r >> 1;
  if (l == r) return;
  DaC(l, mid);
  DaC(mid + 1, r);
  CalcSuf(l, mid);
  CalcPre(mid + 1, r);
  CH.clear();
  int j = l;
  for (int i = r; i > mid; i--) {
    while (j <= mid && suf[j + 1] >= pre[i])
      if (f[t0][j++] != oo) CH.add({suf[j], f[t0][j - 1] - (j - 1) * suf[j]});
    minimize(f[t1][i], CH.query(i));
  }
  CH.clear();
  j = mid;
  for (int i = mid + 1; i <= r; i++) {
    while (j >= l && suf[j + 1] <= pre[i])
      if (f[t0][j--] != oo) CH.add({j + 1, f[t0][j + 1]});
    minimize(f[t1][i], CH.query(-pre[i]) + i * pre[i]);
  }
}
int main() {
  if (fopen("exam.inp", "r"))
    freopen("exam.inp", "r", stdin), freopen("exam.out", "w", stdout);
  int maxx = 0;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    maximize(maxx, a[i]);
    f[t0][i] = maxx * i;
  }
  for (int i = 2; i <= m; i++) {
    fill(f[t1] + 1, f[t1] + 1 + n, oo);
    DaC(1, n);
    t0 ^= 1;
    t1 ^= 1;
  }
  printf("%d", f[t0][n]);
}
