868_F. Yet Another Minimization Problem
You are given an array of n integers a1... an. The cost of a subsegment is the number of unordered pairs of distinct indices within the subsegment that contain equal elements. Split the given array into k non-intersecting non-empty subsegments so that the sum of their costs is minimum possible. Each element should be present in exactly one subsegment.

Input

The first line contains two integers n and k (2 ≤ n ≤ 105, 2 ≤ k ≤ min (n, 20)) — the length of the array and the number of segments you need to split the array into.

The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) — the elements of the array.

Output

Print single integer: the minimum possible total cost of resulting subsegments.

Examples

Input

7 3
1 1 3 3 3 2 1


Output

1


Input

10 2
1 2 1 2 1 2 1 2 1 2


Output

8


Input

13 3
1 2 2 2 1 2 1 1 1 2 2 1 1


Output

9

Note

In the first example it's optimal to split the sequence into the following three subsegments: [1], [1, 3], [3, 3, 2, 1]. The costs are 0, 0 and 1, thus the answer is 1.

In the second example it's optimal to split the sequence in two equal halves. The cost for each half is 4.

In the third example it's optimal to split the sequence in the following way: [1, 2, 2, 2, 1], [2, 1, 1, 1, 2], [2, 1, 1]. The costs are 4, 4, 1.
```python
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
a = [0]+list(map(int, input().split()))
buc = [0]*(n+1)
dp_p = [n*(n-1)//2]*(n+1)
dp_c = [0]*(n+1)
dp_p[0] = 0
buc[a[1]] = 1
L = R = 1
ans = 0


def cal(l, r):
    global L, R, ans
    while L < l:
        ans += 1-buc[a[L]]
        buc[a[L]] -= 1
        L += 1
    while L > l:
        L -= 1
        ans += buc[a[L]]
        buc[a[L]] += 1
    while R < r:
        R += 1
        ans += buc[a[R]]
        buc[a[R]] += 1
    while R > r:
        ans += 1-buc[a[R]]
        buc[a[R]] -= 1
        R -= 1


def solve(lb, rb, l, r):
    global ans, L
    if lb > rb or l > r:
        return
    mid = (l+r)//2
    d = 0
    res = 9223372036854775807
    cal(lb, mid)
    for i in range(lb, rb+1):
        ans += 1-buc[a[L]]
        buc[a[L]] -= 1
        L += 1
        if res > dp_p[i]+ans:
            res = dp_p[i]+ans
            d = i
    dp_c[mid] = res
    solve(lb, d, l, mid-1)
    solve(d, rb, mid+1, r)


for cur in range(1, m+1):
    solve(0, n-1, 1, n)
    dp_p, dp_c = dp_c, dp_p
print(dp_p[n])

```