#include <bits/stdc++.h>
using namespace std;
long long a[1000005];
struct sgt {
  long long t[1000005 << 2], id[1000005 << 2];
  sgt() { memset(t, 0, sizeof(t)); }
  void pushup(int rt) {
    if (t[rt << 1] > t[rt << 1 | 1])
      t[rt] = t[rt << 1], id[rt] = id[rt << 1];
    else
      t[rt] = t[rt << 1 | 1], id[rt] = id[rt << 1 | 1];
  }
  void build(int l, int r, int rt) {
    if (l == r) {
      t[rt] = a[l];
      id[rt] = l;
      return;
    }
    int mid = l + r >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    pushup(rt);
  }
  pair<long long, long long> query(int l, int r, int rt, int ll, int rr) {
    if (ll <= l && rr >= r) return pair<long long, long long>(t[rt], id[rt]);
    int mid = l + r >> 1;
    pair<long long, long long> ans = pair<long long, long long>(0, 0);
    if (ll <= mid) {
      pair<long long, long long> p = query(l, mid, rt << 1, ll, rr);
      if (p.first > ans.first) ans.first = p.first, ans.second = p.second;
    }
    if (rr > mid) {
      pair<long long, long long> p = query(mid + 1, r, rt << 1 | 1, ll, rr);
      if (p.first > ans.first) ans.first = p.first, ans.second = p.second;
    }
    return ans;
  }
} t;
long long n, k;
long long ans;
const long long mo = 1e9 + 7;
long long calc(long long &x, long long y, long long d) {
  long long num = (y - x) / d + 1;
  y = (num - 1) * d + x;
  long long xx = x;
  x = y;
  return (xx + y) * num / 2;
}
void solve(long long l, long long r) {
  if (l + k - 1 > r) return;
  pair<long long, long long> p = t.query(0, 1000005, 1, l, r);
  long long s = p.first, pos = p.second;
  long long len = r - l + 1;
  long long x = pos - l + 1, y = r - pos + 1;
  if (x > y) swap(x, y);
  long long kk = k;
  long long sum = 0;
  if (x >= kk) sum += calc(kk, x, k - 1), kk += k - 1;
  long long q = (y - kk) / (k - 1);
  if (y >= kk) {
    sum += (q + 1) * x;
    kk += (q + 1) * (k - 1);
  }
  if (len >= kk) {
    long long t = len - kk + 1;
    sum += calc(t, 0, -k + 1);
  }
  sum %= mo;
  ans += (1ll * sum * s) % mo;
  ans %= mo;
  solve(l, pos - 1);
  solve(pos + 1, r);
}
int main() {
  scanf("%lld%lld", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  t.build(0, 1000005, 1);
  solve(1, n);
  printf("%lld\n", ans);
}

### Prompt

Create a solution in Cpp for the following problem:

Given an array a of n integers and an integer k (2 ≤ k ≤ n), where each element of the array is denoted by a_i (0 ≤ i < n). Perform the operation z given below on a and print the value of z(a,k) modulo 10^{9}+7.
    
    
      
    function z(array a, integer k):  
        if length(a) < k:  
            return 0  
        else:  
            b = empty array  
            ans = 0  
            for i = 0 .. (length(a) - k):  
                temp = a[i]  
                for j = i .. (i + k - 1):  
                    temp = max(temp, a[j])  
                append temp to the end of b  
                ans = ans + temp  
            return ans + z(b, k)  
    

Input

The first line of input contains two integers n and k (2 ≤ k ≤ n ≤ 10^6) — the length of the initial array a and the parameter k.

The second line of input contains n integers a_0, a_1, …, a_{n - 1} (1 ≤ a_{i} ≤ 10^9) — the elements of the array a.

Output

Output the only integer, the value of z(a,k) modulo 10^9+7.

Examples

Input

3 2
9 1 10


Output

29


Input

5 3
5 8 7 1 9


Output

34

Note

In the first example: 

  * for a=(9,1,10), ans=19 and b=(9,10), 
  * for a=(9,10), ans=10 and b=(10), 
  * for a=(10), ans=0. 



So the returned value is 19+10+0=29.

In the second example: 

  * for a=(5,8,7,1,9), ans=25 and b=(8,8,9), 
  * for a=(8,8,9), ans=9 and b=(9), 
  * for a=(9), ans=0. 



So the returned value is 25+9+0=34.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long a[1000005];
struct sgt {
  long long t[1000005 << 2], id[1000005 << 2];
  sgt() { memset(t, 0, sizeof(t)); }
  void pushup(int rt) {
    if (t[rt << 1] > t[rt << 1 | 1])
      t[rt] = t[rt << 1], id[rt] = id[rt << 1];
    else
      t[rt] = t[rt << 1 | 1], id[rt] = id[rt << 1 | 1];
  }
  void build(int l, int r, int rt) {
    if (l == r) {
      t[rt] = a[l];
      id[rt] = l;
      return;
    }
    int mid = l + r >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    pushup(rt);
  }
  pair<long long, long long> query(int l, int r, int rt, int ll, int rr) {
    if (ll <= l && rr >= r) return pair<long long, long long>(t[rt], id[rt]);
    int mid = l + r >> 1;
    pair<long long, long long> ans = pair<long long, long long>(0, 0);
    if (ll <= mid) {
      pair<long long, long long> p = query(l, mid, rt << 1, ll, rr);
      if (p.first > ans.first) ans.first = p.first, ans.second = p.second;
    }
    if (rr > mid) {
      pair<long long, long long> p = query(mid + 1, r, rt << 1 | 1, ll, rr);
      if (p.first > ans.first) ans.first = p.first, ans.second = p.second;
    }
    return ans;
  }
} t;
long long n, k;
long long ans;
const long long mo = 1e9 + 7;
long long calc(long long &x, long long y, long long d) {
  long long num = (y - x) / d + 1;
  y = (num - 1) * d + x;
  long long xx = x;
  x = y;
  return (xx + y) * num / 2;
}
void solve(long long l, long long r) {
  if (l + k - 1 > r) return;
  pair<long long, long long> p = t.query(0, 1000005, 1, l, r);
  long long s = p.first, pos = p.second;
  long long len = r - l + 1;
  long long x = pos - l + 1, y = r - pos + 1;
  if (x > y) swap(x, y);
  long long kk = k;
  long long sum = 0;
  if (x >= kk) sum += calc(kk, x, k - 1), kk += k - 1;
  long long q = (y - kk) / (k - 1);
  if (y >= kk) {
    sum += (q + 1) * x;
    kk += (q + 1) * (k - 1);
  }
  if (len >= kk) {
    long long t = len - kk + 1;
    sum += calc(t, 0, -k + 1);
  }
  sum %= mo;
  ans += (1ll * sum * s) % mo;
  ans %= mo;
  solve(l, pos - 1);
  solve(pos + 1, r);
}
int main() {
  scanf("%lld%lld", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  t.build(0, 1000005, 1);
  solve(1, n);
  printf("%lld\n", ans);
}

```