#include <bits/stdc++.h>
using namespace std;
long long inv[200005], f[200005], jie[200005];
long long qmi(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = ans * a % 1000000007;
    a = a * a % 1000000007;
    b /= 2;
  }
  return ans;
}
long long C(long long x, long long y) {
  long long ans;
  ans = jie[x] * f[y] % 1000000007 * f[x - y] % 1000000007;
  return ans;
}
long long S(long long n, long long k) {
  long long ans = 0;
  for (int i = 0; i <= k; i++) {
    if ((k + i) % 2)
      ans =
          ((ans - C(k, i) * qmi(i, n) % 1000000007) % 1000000007 + 1000000007) %
          1000000007;
    else
      ans =
          ((ans + C(k, i) * qmi(i, n) % 1000000007) % 1000000007 + 1000000007) %
          1000000007;
  }
  ans = ans * f[k] % 1000000007;
  return ans;
}
int main() {
  jie[0] = f[0] = inv[0] = 1;
  for (long long i = 1; i < 200005; i++) inv[i] = qmi(i, 1000000007 - 2);
  for (long long i = 1; i < 200005; i++)
    f[i] = f[i - 1] * inv[i] % 1000000007, jie[i] = jie[i - 1] * i % 1000000007;
  long long n, k, sum = 0;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    long long r;
    cin >> r;
    sum = (sum + r) % 1000000007;
  }
  long long res = (S(n, k) + (n - 1) * S(n - 1, k) % 1000000007) % 1000000007 *
                  sum % 1000000007;
  cout << res << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
long long inv[200005], f[200005], jie[200005];
long long qmi(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = ans * a % 1000000007;
    a = a * a % 1000000007;
    b /= 2;
  }
  return ans;
}
long long C(long long x, long long y) {
  long long ans;
  ans = jie[x] * f[y] % 1000000007 * f[x - y] % 1000000007;
  return ans;
}
long long S(long long n, long long k) {
  long long ans = 0;
  for (int i = 0; i <= k; i++) {
    if ((k + i) % 2)
      ans =
          ((ans - C(k, i) * qmi(i, n) % 1000000007) % 1000000007 + 1000000007) %
          1000000007;
    else
      ans =
          ((ans + C(k, i) * qmi(i, n) % 1000000007) % 1000000007 + 1000000007) %
          1000000007;
  }
  ans = ans * f[k] % 1000000007;
  return ans;
}
int main() {
  jie[0] = f[0] = inv[0] = 1;
  for (long long i = 1; i < 200005; i++) inv[i] = qmi(i, 1000000007 - 2);
  for (long long i = 1; i < 200005; i++)
    f[i] = f[i - 1] * inv[i] % 1000000007, jie[i] = jie[i - 1] * i % 1000000007;
  long long n, k, sum = 0;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    long long r;
    cin >> r;
    sum = (sum + r) % 1000000007;
  }
  long long res = (S(n, k) + (n - 1) * S(n - 1, k) % 1000000007) % 1000000007 *
                  sum % 1000000007;
  cout << res << endl;
}

```