#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, skc = 1e9 + 7;
int n, k, fac[N], inv[N], a[N];
int C(int A, int B) { return 1ll * fac[B] * inv[A] % skc * inv[B - A] % skc; }
void upd(int &x, int y) { x = (x + y) % skc; }
int qpow(int A, int B) {
  int C = 1;
  for (; B; B >>= 1, A = 1ll * A * A % skc)
    if (B & 1) C = 1ll * A * C % skc;
  return C;
}
void prep() {
  fac[0] = inv[0] = 1;
  for (register int i = (1); i <= (n); ++i) fac[i] = 1ll * i * fac[i - 1] % skc;
  inv[n] = qpow(fac[n], skc - 2);
  for (register int i = (n - 1); i >= (1); --i)
    inv[i] = 1ll * (i + 1) * inv[i + 1] % skc;
}
int S(int A, int B) {
  if (A > B) return 0;
  int ans = 0;
  long long nw = 1;
  for (register int i = (0); i <= (A); ++i)
    upd(ans, nw * C(i, A) * qpow(A - i, B) % skc), nw = -nw;
  ans = 1ll * ans * inv[A] % skc;
  return (ans + skc) % skc;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  prep();
  int ans = 0, x;
  int t = S(k, n) + 1ll * S(k, n - 1) * (n - 1) % skc;
  t %= skc;
  for (register int i = (1); i <= (n); ++i)
    cin >> x, upd(ans, 1ll * x * t % skc);
  cout << ans << '\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;
const int N = 2e5 + 10, skc = 1e9 + 7;
int n, k, fac[N], inv[N], a[N];
int C(int A, int B) { return 1ll * fac[B] * inv[A] % skc * inv[B - A] % skc; }
void upd(int &x, int y) { x = (x + y) % skc; }
int qpow(int A, int B) {
  int C = 1;
  for (; B; B >>= 1, A = 1ll * A * A % skc)
    if (B & 1) C = 1ll * A * C % skc;
  return C;
}
void prep() {
  fac[0] = inv[0] = 1;
  for (register int i = (1); i <= (n); ++i) fac[i] = 1ll * i * fac[i - 1] % skc;
  inv[n] = qpow(fac[n], skc - 2);
  for (register int i = (n - 1); i >= (1); --i)
    inv[i] = 1ll * (i + 1) * inv[i + 1] % skc;
}
int S(int A, int B) {
  if (A > B) return 0;
  int ans = 0;
  long long nw = 1;
  for (register int i = (0); i <= (A); ++i)
    upd(ans, nw * C(i, A) * qpow(A - i, B) % skc), nw = -nw;
  ans = 1ll * ans * inv[A] % skc;
  return (ans + skc) % skc;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  prep();
  int ans = 0, x;
  int t = S(k, n) + 1ll * S(k, n - 1) * (n - 1) % skc;
  t %= skc;
  for (register int i = (1); i <= (n); ++i)
    cin >> x, upd(ans, 1ll * x * t % skc);
  cout << ans << '\n';
  return 0;
}

```