#include <bits/stdc++.h>
using namespace std;
template <class T>
bool ckmax(T& x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <class T>
bool ckmin(T& x, T y) {
  return x > y ? x = y, 1 : 0;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? x : -x;
}
const int N = 200005;
inline int qpow(int n, int k) {
  if (k < 0) return 0;
  int res = 1;
  for (; k; k >>= 1, n = 1ll * n * n % 1000000007)
    if (k & 1) res = 1ll * n * res % 1000000007;
  return res;
}
int n, k, sum, fac[N], ifc[N], ans;
int comb(int n, int m) {
  return n < m ? 0
               : 1ll * fac[n] * ifc[m] % 1000000007 * ifc[n - m] % 1000000007;
}
signed main() {
  n = read(), k = read() - 1;
  for (int i = 1, iend = n; i <= iend; ++i) sum = (sum + read()) % 1000000007;
  fac[0] = 1;
  for (int i = 1, iend = k; i <= iend; ++i)
    fac[i] = 1ll * i * fac[i - 1] % 1000000007;
  ifc[k] = qpow(fac[k], 1000000007 - 2);
  for (int i = k - 1, iend = 0; i >= iend; --i)
    ifc[i] = 1ll * ifc[i + 1] * (i + 1) % 1000000007;
  for (int i = 0; i <= k; ++i) {
    int tmp =
        1ll * comb(k, i) *
        (qpow(1 + i, n - 1) + 1ll * (n - 1) * qpow(1 + i, n - 2) % 1000000007) %
        1000000007;
    (k - i)& 1 ? ans = (ans + 1000000007 - tmp) % 1000000007
               : ans = (ans + tmp) % 1000000007;
  }
  ans = 1ll * ans * ifc[k] % 1000000007 * sum % 1000000007;
  cout << ans << '\n';
}

### Prompt

Please provide a CPP coded solution 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;
template <class T>
bool ckmax(T& x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <class T>
bool ckmin(T& x, T y) {
  return x > y ? x = y, 1 : 0;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? x : -x;
}
const int N = 200005;
inline int qpow(int n, int k) {
  if (k < 0) return 0;
  int res = 1;
  for (; k; k >>= 1, n = 1ll * n * n % 1000000007)
    if (k & 1) res = 1ll * n * res % 1000000007;
  return res;
}
int n, k, sum, fac[N], ifc[N], ans;
int comb(int n, int m) {
  return n < m ? 0
               : 1ll * fac[n] * ifc[m] % 1000000007 * ifc[n - m] % 1000000007;
}
signed main() {
  n = read(), k = read() - 1;
  for (int i = 1, iend = n; i <= iend; ++i) sum = (sum + read()) % 1000000007;
  fac[0] = 1;
  for (int i = 1, iend = k; i <= iend; ++i)
    fac[i] = 1ll * i * fac[i - 1] % 1000000007;
  ifc[k] = qpow(fac[k], 1000000007 - 2);
  for (int i = k - 1, iend = 0; i >= iend; --i)
    ifc[i] = 1ll * ifc[i + 1] * (i + 1) % 1000000007;
  for (int i = 0; i <= k; ++i) {
    int tmp =
        1ll * comb(k, i) *
        (qpow(1 + i, n - 1) + 1ll * (n - 1) * qpow(1 + i, n - 2) % 1000000007) %
        1000000007;
    (k - i)& 1 ? ans = (ans + 1000000007 - tmp) % 1000000007
               : ans = (ans + tmp) % 1000000007;
  }
  ans = 1ll * ans * ifc[k] % 1000000007 * sum % 1000000007;
  cout << ans << '\n';
}

```