#include <bits/stdc++.h>
using namespace std;
inline int gi() {
  register int data = 0, w = 1;
  register char ch = 0;
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
  return w * data;
}
const int MAX_N = 2e5 + 5, Mod = 1e9 + 7;
int fpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % Mod;
    x = 1ll * x * x % Mod;
    y >>= 1;
  }
  return res;
}
int fac[MAX_N], ifc[MAX_N];
int C(int n, int m) {
  if (n < m)
    return 0;
  else
    return 1ll * fac[n] * ifc[n - m] % Mod * ifc[m] % Mod;
}
int N, K, w[MAX_N];
int S(int n, int m) {
  int res = 0, p = 1;
  for (int i = 0; i <= m; i++) {
    res = (res + 1ll * p * C(m, i) % Mod * fpow(m - i, n)) % Mod;
    p = Mod - p;
  }
  res = 1ll * res * ifc[m] % Mod;
  return res;
}
int main() {
  N = gi(), K = gi();
  for (int i = 1; i <= N; i++) w[i] = gi();
  fac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = 1ll * i * fac[i - 1] % Mod;
  ifc[N] = fpow(fac[N], Mod - 2);
  for (int i = N - 1; ~i; i--) ifc[i] = 1ll * ifc[i + 1] * (i + 1) % Mod;
  int ans = 0;
  for (int i = 1; i <= N; i++) ans = (ans + w[i]) % Mod;
  printf("%lld\n",
         1ll * ans * (S(N, K) + 1ll * S(N - 1, K) * (N - 1) % 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>
using namespace std;
inline int gi() {
  register int data = 0, w = 1;
  register char ch = 0;
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
  return w * data;
}
const int MAX_N = 2e5 + 5, Mod = 1e9 + 7;
int fpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % Mod;
    x = 1ll * x * x % Mod;
    y >>= 1;
  }
  return res;
}
int fac[MAX_N], ifc[MAX_N];
int C(int n, int m) {
  if (n < m)
    return 0;
  else
    return 1ll * fac[n] * ifc[n - m] % Mod * ifc[m] % Mod;
}
int N, K, w[MAX_N];
int S(int n, int m) {
  int res = 0, p = 1;
  for (int i = 0; i <= m; i++) {
    res = (res + 1ll * p * C(m, i) % Mod * fpow(m - i, n)) % Mod;
    p = Mod - p;
  }
  res = 1ll * res * ifc[m] % Mod;
  return res;
}
int main() {
  N = gi(), K = gi();
  for (int i = 1; i <= N; i++) w[i] = gi();
  fac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = 1ll * i * fac[i - 1] % Mod;
  ifc[N] = fpow(fac[N], Mod - 2);
  for (int i = N - 1; ~i; i--) ifc[i] = 1ll * ifc[i + 1] * (i + 1) % Mod;
  int ans = 0;
  for (int i = 1; i <= N; i++) ans = (ans + w[i]) % Mod;
  printf("%lld\n",
         1ll * ans * (S(N, K) + 1ll * S(N - 1, K) * (N - 1) % Mod) % Mod);
  return 0;
}

```