#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int n, k, sum, ans, inv[200005], facinv[200005];
inline int GetInt() {
  int num = 0, k = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') k = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = num * 10 + ch - '0';
    ch = getchar();
  }
  return num * k;
}
inline int PlusMod(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int MinusMod(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline int QuickPower(int x, int k) {
  int s = 1;
  if (k < 0) return 1;
  while (k) {
    if (k & 1) s = 1ll * s * x % mod;
    k >>= 1;
    x = 1ll * x * x % mod;
  }
  return s;
}
int main() {
  n = GetInt(), k = GetInt();
  for (int i = 1; i <= n; i++) sum = PlusMod(sum, GetInt());
  facinv[0] = facinv[1] = inv[1] = 1;
  for (int i = 2; i <= k; i++) {
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    facinv[i] = 1ll * facinv[i - 1] * inv[i] % mod;
  }
  for (int i = 0; i < k; i++) {
    int tmp = (i & 1 ? -1ll : 1ll) * facinv[i] * facinv[k - 1 - i] % mod;
    ans = (1ll * ans +
           1ll * tmp * QuickPower(k - i, n - 2) % mod * (k - i + n - 1) % mod) %
          mod;
  }
  ans = (ans + mod) % mod;
  ans = 1ll * ans * sum % mod;
  printf("%d", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 mod = 1e9 + 7;
int n, k, sum, ans, inv[200005], facinv[200005];
inline int GetInt() {
  int num = 0, k = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') k = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = num * 10 + ch - '0';
    ch = getchar();
  }
  return num * k;
}
inline int PlusMod(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int MinusMod(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline int QuickPower(int x, int k) {
  int s = 1;
  if (k < 0) return 1;
  while (k) {
    if (k & 1) s = 1ll * s * x % mod;
    k >>= 1;
    x = 1ll * x * x % mod;
  }
  return s;
}
int main() {
  n = GetInt(), k = GetInt();
  for (int i = 1; i <= n; i++) sum = PlusMod(sum, GetInt());
  facinv[0] = facinv[1] = inv[1] = 1;
  for (int i = 2; i <= k; i++) {
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    facinv[i] = 1ll * facinv[i - 1] * inv[i] % mod;
  }
  for (int i = 0; i < k; i++) {
    int tmp = (i & 1 ? -1ll : 1ll) * facinv[i] * facinv[k - 1 - i] % mod;
    ans = (1ll * ans +
           1ll * tmp * QuickPower(k - i, n - 2) % mod * (k - i + n - 1) % mod) %
          mod;
  }
  ans = (ans + mod) % mod;
  ans = 1ll * ans * sum % mod;
  printf("%d", ans);
  return 0;
}

```