#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void chkmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void chkmin(T &a, T b) {
  if (a > b) a = b;
}
inline int read() {
  int sum = 0, p = 1;
  char ch = getchar();
  while (!(('0' <= ch && ch <= '9') || ch == '-')) ch = getchar();
  if (ch == '-') p = -1, ch = getchar();
  while ('0' <= ch && ch <= '9') sum = sum * 10 + ch - 48, ch = getchar();
  return sum * p;
}
const int mod = 1e9 + 7;
const int maxn = 2e5 + 20;
inline int power(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (long long)ans * a % mod;
    b >>= 1;
    a = (long long)a * a % mod;
  }
  return ans;
}
int n, k;
int W;
int jc[maxn], jcn[maxn], inv[maxn];
inline void init() {
  n = read();
  k = read();
  for (int i = (1), _end_ = (n); i <= _end_; i++) W = (W + read()) % mod;
  jc[0] = jc[1] = jcn[0] = jcn[1] = inv[1] = 1;
  for (int i = (2), _end_ = (n); i <= _end_; i++)
    jc[i] = (long long)i * jc[i - 1] % mod,
    inv[i] = (long long)(mod - mod / i) * inv[mod % i] % mod,
    jcn[i] = (long long)jcn[i - 1] * inv[i] % mod;
}
inline int S(int n, int m) {
  int ans = 0;
  for (int i = (1), _end_ = (m); i <= _end_; i++)
    ans = (ans + (long long)(m - i & 1 ? -1 : 1) * jcn[i] * jcn[m - i] % mod *
                     power(i, n)) %
          mod;
  return (ans + mod) % mod;
}
inline void doing() {
  int res = (S(n, k) + (long long)(n - 1) * S(n - 1, k)) % mod;
  printf("%lld\n", (long long)W * res % mod);
}
int main() {
  init();
  doing();
  return 0;
}

### Prompt

Your task is to create 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;
template <class T>
inline void chkmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void chkmin(T &a, T b) {
  if (a > b) a = b;
}
inline int read() {
  int sum = 0, p = 1;
  char ch = getchar();
  while (!(('0' <= ch && ch <= '9') || ch == '-')) ch = getchar();
  if (ch == '-') p = -1, ch = getchar();
  while ('0' <= ch && ch <= '9') sum = sum * 10 + ch - 48, ch = getchar();
  return sum * p;
}
const int mod = 1e9 + 7;
const int maxn = 2e5 + 20;
inline int power(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (long long)ans * a % mod;
    b >>= 1;
    a = (long long)a * a % mod;
  }
  return ans;
}
int n, k;
int W;
int jc[maxn], jcn[maxn], inv[maxn];
inline void init() {
  n = read();
  k = read();
  for (int i = (1), _end_ = (n); i <= _end_; i++) W = (W + read()) % mod;
  jc[0] = jc[1] = jcn[0] = jcn[1] = inv[1] = 1;
  for (int i = (2), _end_ = (n); i <= _end_; i++)
    jc[i] = (long long)i * jc[i - 1] % mod,
    inv[i] = (long long)(mod - mod / i) * inv[mod % i] % mod,
    jcn[i] = (long long)jcn[i - 1] * inv[i] % mod;
}
inline int S(int n, int m) {
  int ans = 0;
  for (int i = (1), _end_ = (m); i <= _end_; i++)
    ans = (ans + (long long)(m - i & 1 ? -1 : 1) * jcn[i] * jcn[m - i] % mod *
                     power(i, n)) %
          mod;
  return (ans + mod) % mod;
}
inline void doing() {
  int res = (S(n, k) + (long long)(n - 1) * S(n - 1, k)) % mod;
  printf("%lld\n", (long long)W * res % mod);
}
int main() {
  init();
  doing();
  return 0;
}

```