#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
int n, k;
int step;
int LEAFOFFSET;
pair<int, int> IT[4000111];
int modval[1000111];
pair<int, int> MAX(pair<int, int> a, pair<int, int> b) {
  if (a.first > b.first)
    return a;
  else
    return b;
}
pair<int, int> getMax(int L, int R) {
  L += LEAFOFFSET;
  R += LEAFOFFSET;
  pair<int, int> ans = MAX(IT[L], IT[R]);
  while (R - L > 1) {
    if (L % 2 == 0) ans = MAX(ans, IT[L + 1]);
    if (R % 2 == 1) ans = MAX(ans, IT[R - 1]);
    L /= 2;
    R /= 2;
  }
  return ans;
}
vector<pair<pair<int, int>, int> > seqA, seqB;
void clearSeq() {
  seqA.clear();
  seqB.clear();
}
void addSeq(int l, int r) {
  int addMore = step - modval[l];
  if (addMore == step) addMore = 0;
  if (l + addMore <= r) {
    if (addMore != 0) {
      seqA.push_back(make_pair(make_pair(modval[l], step - 1), 1));
    }
    l += addMore;
    int dist = r - l + 1;
    int fullHits = dist / step;
    seqA.push_back(make_pair(make_pair(0, step - 1), fullHits));
    if (modval[r] != step - 1) {
      seqA.push_back(make_pair(make_pair(0, modval[r]), 1));
    }
  } else {
    seqA.push_back(make_pair(make_pair(modval[l], modval[r]), 1));
  }
}
void logSeq() {
  int i;
  for (i = 0; i < seqA.size(); i++) {
    seqB.push_back(seqA[i]);
  }
  seqA.clear();
  return;
}
long long countSeq() {
  int i, j;
  int l, r;
  long long ans = 0, val;
  for (i = 0; i < seqA.size(); i++) {
    for (j = 0; j < seqB.size(); j++) {
      l = seqA[i].first.first;
      if (seqB[j].first.first > l) l = seqB[j].first.first;
      r = seqA[i].first.second;
      if (seqB[j].first.second < r) r = seqB[j].first.second;
      if (r >= l) {
        val = (long long)(r - l + 1) * (long long)seqA[i].second;
        if (val >= MOD) val %= MOD;
        val *= (long long)seqB[j].second;
        ans += val;
        if (ans >= MOD) ans %= MOD;
      }
    }
  }
  return ans;
}
long long solve(int L, int R) {
  if (L >= R) return 0LL;
  pair<int, int> res = getMax(L, R);
  int ind = res.second;
  long long ans = solve(L, ind - 1) + solve(ind + 1, R);
  int i;
  clearSeq();
  addSeq(L, ind);
  logSeq();
  addSeq(ind, R);
  long long times = countSeq() - 1;
  if (times < 0) times += MOD;
  ans += times * (long long)res.first;
  return ans % MOD;
}
int main() {
  int i, j;
  scanf("%d %d", &n, &k);
  step = k - 1;
  LEAFOFFSET = 1;
  while (LEAFOFFSET < n) LEAFOFFSET *= 2;
  LEAFOFFSET--;
  modval[0] = 0;
  for (i = 1; i <= n; i++) {
    modval[i] = modval[i - 1] + 1;
    if (modval[i] == step) modval[i] = 0;
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &IT[i + LEAFOFFSET].first);
    IT[i + LEAFOFFSET].second = i;
  }
  for (i = LEAFOFFSET; i >= 1; i--) {
    IT[i] = MAX(IT[2 * i], IT[2 * i + 1]);
  }
  printf("%lld\n", solve(1, n));
  return 0;
}

### Prompt

Please formulate a cpp solution to 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;
const long long MOD = 1000000007;
int n, k;
int step;
int LEAFOFFSET;
pair<int, int> IT[4000111];
int modval[1000111];
pair<int, int> MAX(pair<int, int> a, pair<int, int> b) {
  if (a.first > b.first)
    return a;
  else
    return b;
}
pair<int, int> getMax(int L, int R) {
  L += LEAFOFFSET;
  R += LEAFOFFSET;
  pair<int, int> ans = MAX(IT[L], IT[R]);
  while (R - L > 1) {
    if (L % 2 == 0) ans = MAX(ans, IT[L + 1]);
    if (R % 2 == 1) ans = MAX(ans, IT[R - 1]);
    L /= 2;
    R /= 2;
  }
  return ans;
}
vector<pair<pair<int, int>, int> > seqA, seqB;
void clearSeq() {
  seqA.clear();
  seqB.clear();
}
void addSeq(int l, int r) {
  int addMore = step - modval[l];
  if (addMore == step) addMore = 0;
  if (l + addMore <= r) {
    if (addMore != 0) {
      seqA.push_back(make_pair(make_pair(modval[l], step - 1), 1));
    }
    l += addMore;
    int dist = r - l + 1;
    int fullHits = dist / step;
    seqA.push_back(make_pair(make_pair(0, step - 1), fullHits));
    if (modval[r] != step - 1) {
      seqA.push_back(make_pair(make_pair(0, modval[r]), 1));
    }
  } else {
    seqA.push_back(make_pair(make_pair(modval[l], modval[r]), 1));
  }
}
void logSeq() {
  int i;
  for (i = 0; i < seqA.size(); i++) {
    seqB.push_back(seqA[i]);
  }
  seqA.clear();
  return;
}
long long countSeq() {
  int i, j;
  int l, r;
  long long ans = 0, val;
  for (i = 0; i < seqA.size(); i++) {
    for (j = 0; j < seqB.size(); j++) {
      l = seqA[i].first.first;
      if (seqB[j].first.first > l) l = seqB[j].first.first;
      r = seqA[i].first.second;
      if (seqB[j].first.second < r) r = seqB[j].first.second;
      if (r >= l) {
        val = (long long)(r - l + 1) * (long long)seqA[i].second;
        if (val >= MOD) val %= MOD;
        val *= (long long)seqB[j].second;
        ans += val;
        if (ans >= MOD) ans %= MOD;
      }
    }
  }
  return ans;
}
long long solve(int L, int R) {
  if (L >= R) return 0LL;
  pair<int, int> res = getMax(L, R);
  int ind = res.second;
  long long ans = solve(L, ind - 1) + solve(ind + 1, R);
  int i;
  clearSeq();
  addSeq(L, ind);
  logSeq();
  addSeq(ind, R);
  long long times = countSeq() - 1;
  if (times < 0) times += MOD;
  ans += times * (long long)res.first;
  return ans % MOD;
}
int main() {
  int i, j;
  scanf("%d %d", &n, &k);
  step = k - 1;
  LEAFOFFSET = 1;
  while (LEAFOFFSET < n) LEAFOFFSET *= 2;
  LEAFOFFSET--;
  modval[0] = 0;
  for (i = 1; i <= n; i++) {
    modval[i] = modval[i - 1] + 1;
    if (modval[i] == step) modval[i] = 0;
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &IT[i + LEAFOFFSET].first);
    IT[i + LEAFOFFSET].second = i;
  }
  for (i = LEAFOFFSET; i >= 1; i--) {
    IT[i] = MAX(IT[2 * i], IT[2 * i + 1]);
  }
  printf("%lld\n", solve(1, n));
  return 0;
}

```