#include <bits/stdc++.h>
using namespace std;
inline int gi() {
  int f = 1, sum = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = (sum << 3) + (sum << 1) + ch - '0';
    ch = getchar();
  }
  return f * sum;
}
const int Mod = 1e9 + 7;
int n, k, fac[200010], fai[200010], inv[200010];
int qpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % Mod;
    b >>= 1;
    a = 1ll * a * a % Mod;
  }
  return ret;
}
int S(int n, int m) {
  int ret = 0;
  for (int i = 0; i <= m; i++)
    ret = (ret + 1ll * (i & 1 ? Mod - 1 : 1) * fai[i] % Mod * fai[m - i] % Mod *
                     qpow(m - i, n) % Mod) %
          Mod;
  return ret;
}
int main() {
  n = gi();
  k = gi();
  int sum = 0;
  inv[0] = inv[1] = fac[0] = fai[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
  for (int i = 2; i <= n; i++)
    inv[i] = 1ll * (Mod - Mod / i) * inv[Mod % i] % Mod;
  for (int i = 1; i <= n; i++) fai[i] = 1ll * fai[i - 1] * inv[i] % Mod;
  for (int i = 1; i <= n; i++) sum = (sum + gi()) % Mod;
  printf("%lld\n",
         1ll * sum * (S(n, k) + 1ll * (n - 1) * S(n - 1, k) % 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>
using namespace std;
inline int gi() {
  int f = 1, sum = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = (sum << 3) + (sum << 1) + ch - '0';
    ch = getchar();
  }
  return f * sum;
}
const int Mod = 1e9 + 7;
int n, k, fac[200010], fai[200010], inv[200010];
int qpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % Mod;
    b >>= 1;
    a = 1ll * a * a % Mod;
  }
  return ret;
}
int S(int n, int m) {
  int ret = 0;
  for (int i = 0; i <= m; i++)
    ret = (ret + 1ll * (i & 1 ? Mod - 1 : 1) * fai[i] % Mod * fai[m - i] % Mod *
                     qpow(m - i, n) % Mod) %
          Mod;
  return ret;
}
int main() {
  n = gi();
  k = gi();
  int sum = 0;
  inv[0] = inv[1] = fac[0] = fai[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
  for (int i = 2; i <= n; i++)
    inv[i] = 1ll * (Mod - Mod / i) * inv[Mod % i] % Mod;
  for (int i = 1; i <= n; i++) fai[i] = 1ll * fai[i - 1] * inv[i] % Mod;
  for (int i = 1; i <= n; i++) sum = (sum + gi()) % Mod;
  printf("%lld\n",
         1ll * sum * (S(n, k) + 1ll * (n - 1) * S(n - 1, k) % Mod) % Mod);
  return 0;
}

```