#include <bits/stdc++.h>
using namespace std;
const int N = 200010, mod = 1000000007;
int qp(int k, int x) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    k >>= 1;
  }
  return res;
}
int n, m, k, ans;
int jc[N], inv[N], bin[N];
void init() {
  jc[0] = inv[0] = 1;
  for (int i = 1; i <= N - 10; i++) jc[i] = 1ll * jc[i - 1] * i % mod;
  inv[N - 10] = qp(mod - 2, jc[N - 10]);
  for (int i = N - 10 - 1; i; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
}
int C(int a, int b) {
  if (a < b) return 0;
  return 1ll * jc[a] * inv[b] % mod * inv[a - b] % mod;
}
int s1, s2, sum;
int main() {
  int u, v;
  scanf("%d%d", &n, &k);
  init();
  for (int i = 1; i <= n; i++) scanf("%d", &s1), sum += s1, sum %= mod;
  s1 = 0;
  for (int i = 0; i <= k; i++)
    s1 = (s1 +
          1ll * ((i & 1) ? mod - 1 : 1) * C(k, i) % mod * qp(n, k - i) % mod) %
         mod,
    s2 = (s2 + 1ll * ((i & 1) ? mod - 1 : 1) * C(k, i) % mod *
                   qp(n - 1, k - i) % mod) %
         mod;
  s1 = 1ll * s1 * inv[k] % mod;
  s2 = 1ll * s2 * inv[k] % mod * (n - 1) % mod;
  printf("%lld", 1ll * (s1 + s2) % mod * sum % 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;
const int N = 200010, mod = 1000000007;
int qp(int k, int x) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    k >>= 1;
  }
  return res;
}
int n, m, k, ans;
int jc[N], inv[N], bin[N];
void init() {
  jc[0] = inv[0] = 1;
  for (int i = 1; i <= N - 10; i++) jc[i] = 1ll * jc[i - 1] * i % mod;
  inv[N - 10] = qp(mod - 2, jc[N - 10]);
  for (int i = N - 10 - 1; i; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
}
int C(int a, int b) {
  if (a < b) return 0;
  return 1ll * jc[a] * inv[b] % mod * inv[a - b] % mod;
}
int s1, s2, sum;
int main() {
  int u, v;
  scanf("%d%d", &n, &k);
  init();
  for (int i = 1; i <= n; i++) scanf("%d", &s1), sum += s1, sum %= mod;
  s1 = 0;
  for (int i = 0; i <= k; i++)
    s1 = (s1 +
          1ll * ((i & 1) ? mod - 1 : 1) * C(k, i) % mod * qp(n, k - i) % mod) %
         mod,
    s2 = (s2 + 1ll * ((i & 1) ? mod - 1 : 1) * C(k, i) % mod *
                   qp(n - 1, k - i) % mod) %
         mod;
  s1 = 1ll * s1 * inv[k] % mod;
  s2 = 1ll * s2 * inv[k] % mod * (n - 1) % mod;
  printf("%lld", 1ll * (s1 + s2) % mod * sum % mod);
  return 0;
}

```