#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int mod = 1e9 + 7;
int n, k;
int fac[200005], inv[200005];
inline int ksm(int x, int k) {
  int s = 1;
  if (k < 0) return 1;
  for (; k; k >>= 1, x = 1ll * x * x % mod)
    if (k & 1) s = 1ll * s * x % mod;
  return s;
}
int main() {
  n = read();
  k = read();
  int sum = 0;
  for (int i = 1; i <= n; i++) sum = (sum + read()) % mod;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  inv[n] = ksm(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  int ans = 0;
  for (int j = 0; j <= k - 1; j++)
    if (j & 1)
      ans =
          (ans - 1ll * inv[j] * inv[k - j - 1] % mod *
                     ((1ll * ksm(k - j, n - 2) * (n - 1 + k - j) % mod) % mod) %
                     mod) %
          mod;
    else
      ans =
          (ans + 1ll * inv[j] * inv[k - j - 1] % mod *
                     ((1ll * ksm(k - j, n - 2) * (n - 1 + k - j) % mod) % mod) %
                     mod) %
          mod;
  ans = (ans + mod) % mod;
  cout << 1ll * ans * sum % mod << "\n";
  return 0;
}

### 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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int mod = 1e9 + 7;
int n, k;
int fac[200005], inv[200005];
inline int ksm(int x, int k) {
  int s = 1;
  if (k < 0) return 1;
  for (; k; k >>= 1, x = 1ll * x * x % mod)
    if (k & 1) s = 1ll * s * x % mod;
  return s;
}
int main() {
  n = read();
  k = read();
  int sum = 0;
  for (int i = 1; i <= n; i++) sum = (sum + read()) % mod;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  inv[n] = ksm(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  int ans = 0;
  for (int j = 0; j <= k - 1; j++)
    if (j & 1)
      ans =
          (ans - 1ll * inv[j] * inv[k - j - 1] % mod *
                     ((1ll * ksm(k - j, n - 2) * (n - 1 + k - j) % mod) % mod) %
                     mod) %
          mod;
    else
      ans =
          (ans + 1ll * inv[j] * inv[k - j - 1] % mod *
                     ((1ll * ksm(k - j, n - 2) * (n - 1 + k - j) % mod) % mod) %
                     mod) %
          mod;
  ans = (ans + mod) % mod;
  cout << 1ll * ans * sum % mod << "\n";
  return 0;
}

```