#include <bits/stdc++.h>
using namespace std;
inline int gi() {
  int data = 0, m = 1;
  char ch = 0;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') {
    m = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    data = (data << 1) + (data << 3) + (ch ^ 48);
    ch = getchar();
  }
  return (m) ? data : -data;
}
int n, K, N, Sum, Ans, IFac[2000005], inv[2000005];
inline int Pow(int ts, int js) {
  int al = 1;
  while (js) {
    if (js & 1) al = 1ll * al * ts % 1000000007;
    ts = 1ll * ts * ts % 1000000007;
    js >>= 1;
  }
  return al;
}
int main() {
  n = gi();
  K = gi();
  int Sum = 0;
  for (int i = 1; i <= n; i++) Sum = (Sum + gi()) % 1000000007;
  if (n == 1 && K == 1) return cout << Sum << endl, 0;
  N = max(N, K);
  inv[0] = inv[1] = IFac[0] = IFac[1] = 1;
  for (int i = 2; i <= N; i++)
    inv[i] =
        1ll * (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007,
    IFac[i] = 1ll * IFac[i - 1] * inv[i] % 1000000007;
  Ans = 0;
  for (int j = 0; j <= K - 1; j++) {
    int ret = 0;
    ret = 1ll * (n - 1) * Pow(K - j, n - 2) % 1000000007;
    ret = (ret + Pow(K - j, n - 1)) % 1000000007;
    Ans = (Ans + 1ll * ((j & 1) ? 1000000007 - 1 : 1) * IFac[j] % 1000000007 *
                     IFac[K - j - 1] % 1000000007 * ret % 1000000007) %
          1000000007;
  }
  Ans = 1ll * Ans * Sum % 1000000007;
  cout << Ans << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
inline int gi() {
  int data = 0, m = 1;
  char ch = 0;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') {
    m = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    data = (data << 1) + (data << 3) + (ch ^ 48);
    ch = getchar();
  }
  return (m) ? data : -data;
}
int n, K, N, Sum, Ans, IFac[2000005], inv[2000005];
inline int Pow(int ts, int js) {
  int al = 1;
  while (js) {
    if (js & 1) al = 1ll * al * ts % 1000000007;
    ts = 1ll * ts * ts % 1000000007;
    js >>= 1;
  }
  return al;
}
int main() {
  n = gi();
  K = gi();
  int Sum = 0;
  for (int i = 1; i <= n; i++) Sum = (Sum + gi()) % 1000000007;
  if (n == 1 && K == 1) return cout << Sum << endl, 0;
  N = max(N, K);
  inv[0] = inv[1] = IFac[0] = IFac[1] = 1;
  for (int i = 2; i <= N; i++)
    inv[i] =
        1ll * (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007,
    IFac[i] = 1ll * IFac[i - 1] * inv[i] % 1000000007;
  Ans = 0;
  for (int j = 0; j <= K - 1; j++) {
    int ret = 0;
    ret = 1ll * (n - 1) * Pow(K - j, n - 2) % 1000000007;
    ret = (ret + Pow(K - j, n - 1)) % 1000000007;
    Ans = (Ans + 1ll * ((j & 1) ? 1000000007 - 1 : 1) * IFac[j] % 1000000007 *
                     IFac[K - j - 1] % 1000000007 * ret % 1000000007) %
          1000000007;
  }
  Ans = 1ll * Ans * Sum % 1000000007;
  cout << Ans << endl;
  return 0;
}

```