#include <bits/stdc++.h>
const int mod = 1000000007;
const int N = 2e5;
using namespace std;
int fac[N], inv[N], n, k, ans;
template <typename T>
inline T read() {
  T x = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * w;
}
int fpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod)
    if (y & 1) res = 1ll * res * x % mod;
  return res;
}
int C(int n, int m) { return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod; }
int S(int n, int m) {
  int res = 0;
  for (int i = 0; i <= m; i++)
    res = (1ll * res + 1ll * (i & 1 ? -1 : 1) * C(m, i) * fpow(m - i, n) % mod +
           mod) %
          mod;
  res = 1ll * res * inv[m] % mod;
  return res;
}
int main() {
  n = read<int>(), k = read<int>();
  for (int i = (fac[0] = 1); i <= k; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  inv[k] = fpow(fac[k], mod - 2);
  for (int i = k - 1; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= n; i++) ans = (ans + read<int>()) % mod;
  printf(
      "%lld\n",
      1ll * ans * ((S(n, k) + 1ll * (n - 1) * S(n - 1, k) % mod) % mod) % mod);
  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>
const int mod = 1000000007;
const int N = 2e5;
using namespace std;
int fac[N], inv[N], n, k, ans;
template <typename T>
inline T read() {
  T x = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * w;
}
int fpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod)
    if (y & 1) res = 1ll * res * x % mod;
  return res;
}
int C(int n, int m) { return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod; }
int S(int n, int m) {
  int res = 0;
  for (int i = 0; i <= m; i++)
    res = (1ll * res + 1ll * (i & 1 ? -1 : 1) * C(m, i) * fpow(m - i, n) % mod +
           mod) %
          mod;
  res = 1ll * res * inv[m] % mod;
  return res;
}
int main() {
  n = read<int>(), k = read<int>();
  for (int i = (fac[0] = 1); i <= k; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  inv[k] = fpow(fac[k], mod - 2);
  for (int i = k - 1; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= n; i++) ans = (ans + read<int>()) % mod;
  printf(
      "%lld\n",
      1ll * ans * ((S(n, k) + 1ll * (n - 1) * S(n - 1, k) % mod) % mod) % mod);
  return 0;
}

```