#include <bits/stdc++.h>
long long gi() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? x : -x;
}
int pow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1ll * ret * x % 1000000007;
    x = 1ll * x * x % 1000000007;
    y >>= 1;
  }
  return ret;
}
int ifact[200010];
int S(int n, int m) {
  int ret = 0;
  for (int i = 0, M = 1; i <= m; ++i, M = 1000000007 - M)
    ret = (ret + 1ll * M * ifact[i] % 1000000007 * ifact[m - i] % 1000000007 *
                     pow(m - i, n)) %
          1000000007;
  return ret;
}
int main() {
  int n = gi(), k = gi(), W = 0;
  for (int i = 1; i <= n; ++i) W = (W + gi()) % 1000000007;
  ifact[0] = 1;
  for (int i = 1; i <= n; ++i) ifact[i] = 1ll * ifact[i - 1] * i % 1000000007;
  ifact[n] = pow(ifact[n], 1000000007 - 2);
  for (int i = n; i; --i) ifact[i - 1] = 1ll * ifact[i] * i % 1000000007;
  printf("%d\n", 1ll * W *
                     (S(n, k) + 1ll * (n - 1) * S(n - 1, k) % 1000000007) %
                     1000000007);
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
long long gi() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? x : -x;
}
int pow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1ll * ret * x % 1000000007;
    x = 1ll * x * x % 1000000007;
    y >>= 1;
  }
  return ret;
}
int ifact[200010];
int S(int n, int m) {
  int ret = 0;
  for (int i = 0, M = 1; i <= m; ++i, M = 1000000007 - M)
    ret = (ret + 1ll * M * ifact[i] % 1000000007 * ifact[m - i] % 1000000007 *
                     pow(m - i, n)) %
          1000000007;
  return ret;
}
int main() {
  int n = gi(), k = gi(), W = 0;
  for (int i = 1; i <= n; ++i) W = (W + gi()) % 1000000007;
  ifact[0] = 1;
  for (int i = 1; i <= n; ++i) ifact[i] = 1ll * ifact[i - 1] * i % 1000000007;
  ifact[n] = pow(ifact[n], 1000000007 - 2);
  for (int i = n; i; --i) ifact[i - 1] = 1ll * ifact[i] * i % 1000000007;
  printf("%d\n", 1ll * W *
                     (S(n, k) + 1ll * (n - 1) * S(n - 1, k) % 1000000007) %
                     1000000007);
  return 0;
}

```