#include <bits/stdc++.h>
inline int in() {
  int k = 0;
  char ch = getchar();
  bool p = 1;
  while (ch < '-') ch = getchar();
  if (ch == '-') ch = getchar(), p = 0;
  while (ch > '-') k = k * 10 + ch - '0', ch = getchar();
  return p ? k : -k;
}
const int YL = 1e9 + 7, N = 2e5 + 5, M = N - 5;
inline int ksm(int a, int k) {
  int r = 1;
  while (k) {
    if (k & 1) r = 1ll * r * a % YL;
    a = 1ll * a * a % YL, k >>= 1;
  }
  return r;
}
inline int MO(const int &x) { return x >= YL ? x - YL : x; }
int fac[N], fnv[N];
inline int C(int x, int y) {
  return y > x ? 0 : 1ll * fac[x] * fnv[y] % YL * fnv[x - y] % YL;
}
inline int S(int n, int m) {
  int ans = 0;
  for (int i = 0, d = 1; i <= m; ++i, d = YL - d)
    ans = (1ll * C(m, i) * d % YL * ksm(m - i, n) + ans) % YL;
  return 1ll * ans * fnv[m] % YL;
}
int main() {
  fac[0] = 1;
  int n = in(), k = in(), sum = 0;
  for (int i = 1; i <= M; ++i) fac[i] = 1ll * fac[i - 1] * i % YL;
  fnv[M] = ksm(fac[M], YL - 2);
  for (int i = M; i >= 1; --i) fnv[i - 1] = 1ll * fnv[i] * i % YL;
  for (int i = 1; i <= n; ++i) sum = MO(sum + in());
  sum = 1ll * sum * ((1ll * S(n - 1, k) * (n + k - 1) + S(n - 1, k - 1)) % YL) %
        YL;
  printf("%d\n", sum);
  return 0;
}

### Prompt

Generate 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>
inline int in() {
  int k = 0;
  char ch = getchar();
  bool p = 1;
  while (ch < '-') ch = getchar();
  if (ch == '-') ch = getchar(), p = 0;
  while (ch > '-') k = k * 10 + ch - '0', ch = getchar();
  return p ? k : -k;
}
const int YL = 1e9 + 7, N = 2e5 + 5, M = N - 5;
inline int ksm(int a, int k) {
  int r = 1;
  while (k) {
    if (k & 1) r = 1ll * r * a % YL;
    a = 1ll * a * a % YL, k >>= 1;
  }
  return r;
}
inline int MO(const int &x) { return x >= YL ? x - YL : x; }
int fac[N], fnv[N];
inline int C(int x, int y) {
  return y > x ? 0 : 1ll * fac[x] * fnv[y] % YL * fnv[x - y] % YL;
}
inline int S(int n, int m) {
  int ans = 0;
  for (int i = 0, d = 1; i <= m; ++i, d = YL - d)
    ans = (1ll * C(m, i) * d % YL * ksm(m - i, n) + ans) % YL;
  return 1ll * ans * fnv[m] % YL;
}
int main() {
  fac[0] = 1;
  int n = in(), k = in(), sum = 0;
  for (int i = 1; i <= M; ++i) fac[i] = 1ll * fac[i - 1] * i % YL;
  fnv[M] = ksm(fac[M], YL - 2);
  for (int i = M; i >= 1; --i) fnv[i - 1] = 1ll * fnv[i] * i % YL;
  for (int i = 1; i <= n; ++i) sum = MO(sum + in());
  sum = 1ll * sum * ((1ll * S(n - 1, k) * (n + k - 1) + S(n - 1, k - 1)) % YL) %
        YL;
  printf("%d\n", sum);
  return 0;
}

```