For a sequence S of positive integers and positive integers k and l, S is said to belong to level (k,l) when one of the following conditions is satisfied:

* The length of S is 1, and its only element is k.
* There exist sequences T_1,T_2,...,T_m (m \geq l) belonging to level (k-1,l) such that the concatenation of T_1,T_2,...,T_m in this order coincides with S.



Note that the second condition has no effect when k=1, that is, a sequence belongs to level (1,l) only if the first condition is satisfied.

Given are a sequence of positive integers A_1,A_2,...,A_N and a positive integer L. Find the number of subsequences A_i,A_{i+1},...,A_j (1 \leq i \leq j \leq N) that satisfy the following condition:

* There exists a positive integer K such that the sequence A_i,A_{i+1},...,A_j belongs to level (K,L).

Constraints

* 1 \leq N \leq 2 \times 10^5
* 2 \leq L \leq N
* 1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N L
A_1 A_2 ... A_N


Output

Print the number of subsequences A_i,A_{i+1},...,A_j (1 \leq i \leq j \leq N) that satisfy the condition.

Examples

Input

9 3
2 1 1 1 1 1 1 2 3


Output

22


Input

9 2
2 1 1 1 1 1 1 2 3


Output

41


Input

15 3
4 3 2 1 1 1 2 3 2 2 1 1 1 2 2


Output

31
#include<bits/stdc++.h>

using namespace std;

#define ll  long long
#define ld  double

#define sz(x)   (int)x.size()
#define all(x)  x.begin(),x.end()

#define pb  emplace_back
#define X   first
#define Y   second

const int   N   = 5e6 + 5;

typedef pair<int,int>   ii;

int a[N], t[N];
int f1[N], f2[N];
int s1[N], s2[N];

int pre[N];
int nxt[N];

int main()  {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    int n;  cin >> n;
    int L;  cin >> L;

    priority_queue<ii,vector<ii>,greater<ii> >  pq;

    for(int i = 1 ; i <= n ; ++i)   {
        cin >> a[i];
        f1[i] = f2[i] = 1;
        pre[i] = i - 1;
        nxt[i] = i + 1;

        pq.push(ii(a[i],i));
    }
    nxt[n] = 0;

    int tot = n;
    ll  ans = 0;

    while (pq.size())   {
        int p = pq.top().Y;
        int v = pq.top().X;

        pq.pop();

        if (t[p])   continue;

        int l = p;
        int r = p;

        while (a[pre[l]] == a[l])   l = pre[l];
        while (a[nxt[r]] == a[r])   r = nxt[r];

        int cnt = 0;
        int prv = 0;

        for(int i = l ; i != nxt[r] ; t[i] = 1, i = nxt[i]) {
            cnt++;

            if (i <= n) prv = cnt;

            s1[cnt] = s1[cnt - 1] + f1[i];
            s2[cnt] = s2[cnt - 1] + f2[i];

            if (prv && cnt >= L)
                ans += 1ll * f2[i] * s1[min(prv,cnt - L + 1)];
        }
        if (cnt < L)    {
            nxt[pre[l]] = 0;
            pre[nxt[r]] = 0;
            continue;
        }
        int S = cnt / L;

        for(int i = 1 ; i < S ; ++i)    {
            nxt[tot + i] = 1 + tot + i;
            pre[tot + i + 1] = tot + i;
        }
        for(int i = 1 ; i <= S ; ++i)
            a[tot + i] = a[p] + 1;

        nxt[pre[l]] = tot + 1;  pre[tot + 1] = pre[l];
        pre[nxt[r]] = tot + S;  nxt[tot + S] = nxt[r];

        for(int i = 1 ; i <= S ; ++i)   {
            int j = S - i + 1;

            f2[tot + i] = s2[min((i + 1) * L - 1,cnt)] - s2[i * L - 1];
            f1[tot + i] = s1[cnt - j * L + 1] - s1[max(cnt - j * L - L + 1,0)];
        }
        pq.push(ii(a[tot + 1],tot + 1));
        tot += S;
    }
    cout << ans + n << endl;
}