#include <bits/stdc++.h>
using namespace std;
int n, k;
int ifac[200005];
int quick_pow(int, int);
int main() {
  scanf("%d%d", &n, &k);
  ifac[0] = 1;
  if (n == 1) {
    int x;
    scanf("%d", &x);
    cout << x << endl;
    return 0;
  }
  for (int i = 1; i <= k; ++i) ifac[i] = 1ll * ifac[i - 1] * i % 1000000007;
  ifac[k] = quick_pow(ifac[k], 1000000007 - 2);
  for (int i = k; i; --i) ifac[i - 1] = 1ll * ifac[i] * i % 1000000007;
  int sum = 0, ans = 0;
  for (int i = 0; i < k; ++i) {
    int tmp1 = 1ll * ifac[i] * ifac[k - 1 - i] % 1000000007;
    int tmp2 = (quick_pow(i + 1, n - 1) +
                1ll * (n - 1) * quick_pow(i + 1, n - 2) % 1000000007) %
               1000000007;
    tmp1 = 1ll * tmp1 * tmp2 % 1000000007;
    if ((k - i - 1) & 1) tmp1 = 1000000007 - tmp1;
    (sum += tmp1) %= 1000000007;
  }
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    (ans += 1ll * x * sum % 1000000007) %= 1000000007;
  }
  cout << ans << endl;
  return 0;
}
int quick_pow(int x, int y) {
  int sum = 1;
  for (; y; y >>= 1, x = 1ll * x * x % 1000000007)
    if (y & 1) sum = 1ll * sum * x % 1000000007;
  return sum;
}

### Prompt

Generate a CPP solution to the following problem:

You are given a set of n elements indexed from 1 to n. The weight of i-th element is wi. The weight of some subset of a given set is denoted as <image>. The weight of some partition R of a given set into k subsets is <image> (recall that a partition of a given set is a set of its subsets such that every element of the given set belongs to exactly one subset in partition).

Calculate the sum of weights of all partitions of a given set into exactly k non-empty subsets, and print it modulo 109 + 7. Two partitions are considered different iff there exist two elements x and y such that they belong to the same set in one of the partitions, and to different sets in another partition.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2·105) — the number of elements and the number of subsets in each partition, respectively.

The second line contains n integers wi (1 ≤ wi ≤ 109)— weights of elements of the set.

Output

Print one integer — the sum of weights of all partitions of a given set into k non-empty subsets, taken modulo 109 + 7.

Examples

Input

4 2
2 3 2 3


Output

160


Input

5 2
1 2 3 4 5


Output

645

Note

Possible partitions in the first sample:

  1. {{1, 2, 3}, {4}}, W(R) = 3·(w1 + w2 + w3) + 1·w4 = 24; 
  2. {{1, 2, 4}, {3}}, W(R) = 26; 
  3. {{1, 3, 4}, {2}}, W(R) = 24; 
  4. {{1, 2}, {3, 4}}, W(R) = 2·(w1 + w2) + 2·(w3 + w4) = 20; 
  5. {{1, 3}, {2, 4}}, W(R) = 20; 
  6. {{1, 4}, {2, 3}}, W(R) = 20; 
  7. {{1}, {2, 3, 4}}, W(R) = 26; 



Possible partitions in the second sample:

  1. {{1, 2, 3, 4}, {5}}, W(R) = 45; 
  2. {{1, 2, 3, 5}, {4}}, W(R) = 48; 
  3. {{1, 2, 4, 5}, {3}}, W(R) = 51; 
  4. {{1, 3, 4, 5}, {2}}, W(R) = 54; 
  5. {{2, 3, 4, 5}, {1}}, W(R) = 57; 
  6. {{1, 2, 3}, {4, 5}}, W(R) = 36; 
  7. {{1, 2, 4}, {3, 5}}, W(R) = 37; 
  8. {{1, 2, 5}, {3, 4}}, W(R) = 38; 
  9. {{1, 3, 4}, {2, 5}}, W(R) = 38; 
  10. {{1, 3, 5}, {2, 4}}, W(R) = 39; 
  11. {{1, 4, 5}, {2, 3}}, W(R) = 40; 
  12. {{2, 3, 4}, {1, 5}}, W(R) = 39; 
  13. {{2, 3, 5}, {1, 4}}, W(R) = 40; 
  14. {{2, 4, 5}, {1, 3}}, W(R) = 41; 
  15. {{3, 4, 5}, {1, 2}}, W(R) = 42. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, k;
int ifac[200005];
int quick_pow(int, int);
int main() {
  scanf("%d%d", &n, &k);
  ifac[0] = 1;
  if (n == 1) {
    int x;
    scanf("%d", &x);
    cout << x << endl;
    return 0;
  }
  for (int i = 1; i <= k; ++i) ifac[i] = 1ll * ifac[i - 1] * i % 1000000007;
  ifac[k] = quick_pow(ifac[k], 1000000007 - 2);
  for (int i = k; i; --i) ifac[i - 1] = 1ll * ifac[i] * i % 1000000007;
  int sum = 0, ans = 0;
  for (int i = 0; i < k; ++i) {
    int tmp1 = 1ll * ifac[i] * ifac[k - 1 - i] % 1000000007;
    int tmp2 = (quick_pow(i + 1, n - 1) +
                1ll * (n - 1) * quick_pow(i + 1, n - 2) % 1000000007) %
               1000000007;
    tmp1 = 1ll * tmp1 * tmp2 % 1000000007;
    if ((k - i - 1) & 1) tmp1 = 1000000007 - tmp1;
    (sum += tmp1) %= 1000000007;
  }
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    (ans += 1ll * x * sum % 1000000007) %= 1000000007;
  }
  cout << ans << endl;
  return 0;
}
int quick_pow(int x, int y) {
  int sum = 1;
  for (; y; y >>= 1, x = 1ll * x * x % 1000000007)
    if (y & 1) sum = 1ll * sum * x % 1000000007;
  return sum;
}

```