#include <bits/stdc++.h>
const int N = 200050;
const int mod = 1000000007;
int F[N], I[N];
int pow(int x, int k) {
  int ret = 1;
  for (; k; k >>= 1, x = (long long)x * x % mod)
    if (k & 1) ret = (long long)ret * x % mod;
  return ret;
}
int get(int n, int k) { return (long long)F[n] * I[k] % mod * I[n - k] % mod; }
int calc(int n, int k) {
  if (k > n) return 0;
  int ret = 0;
  for (int i = k; i; i--) {
    int add = (long long)get(k, i) * pow(i, n) % mod;
    if ((k - i) & 1)
      ret = (ret - add + mod) % mod;
    else
      ret = (ret + add) % mod;
  }
  return (long long)ret * I[k] % mod;
}
int main() {
  int n, k, x, i, sum = 0;
  F[0] = 1;
  for (i = 1; i < N; i++) F[i] = (long long)F[i - 1] * i % mod;
  I[N - 1] = pow(F[N - 1], mod - 2);
  for (i = N - 2; i >= 0; i--) I[i] = (long long)I[i + 1] * (i + 1) % mod;
  scanf("%i %i", &n, &k);
  for (i = 1; i <= n; i++) scanf("%i", &x), sum += x, sum %= mod;
  printf("%i\n", (long long)sum *
                     (calc(n, k) + (long long)(n - 1) * calc(n - 1, k) % mod) %
                     mod);
  return 0;
}

### Prompt

Please formulate 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>
const int N = 200050;
const int mod = 1000000007;
int F[N], I[N];
int pow(int x, int k) {
  int ret = 1;
  for (; k; k >>= 1, x = (long long)x * x % mod)
    if (k & 1) ret = (long long)ret * x % mod;
  return ret;
}
int get(int n, int k) { return (long long)F[n] * I[k] % mod * I[n - k] % mod; }
int calc(int n, int k) {
  if (k > n) return 0;
  int ret = 0;
  for (int i = k; i; i--) {
    int add = (long long)get(k, i) * pow(i, n) % mod;
    if ((k - i) & 1)
      ret = (ret - add + mod) % mod;
    else
      ret = (ret + add) % mod;
  }
  return (long long)ret * I[k] % mod;
}
int main() {
  int n, k, x, i, sum = 0;
  F[0] = 1;
  for (i = 1; i < N; i++) F[i] = (long long)F[i - 1] * i % mod;
  I[N - 1] = pow(F[N - 1], mod - 2);
  for (i = N - 2; i >= 0; i--) I[i] = (long long)I[i + 1] * (i + 1) % mod;
  scanf("%i %i", &n, &k);
  for (i = 1; i <= n; i++) scanf("%i", &x), sum += x, sum %= mod;
  printf("%i\n", (long long)sum *
                     (calc(n, k) + (long long)(n - 1) * calc(n - 1, k) % mod) %
                     mod);
  return 0;
}

```