#include <bits/stdc++.h>
using namespace std;
long long fact[200001], inv_fact[200001];
long long ModExp(int a, int n) {
  long long x = a, y = 1 % 1000000007;
  while (n) {
    if (n % 2) y = (y * x) % 1000000007;
    x = (x * x) % 1000000007;
    n /= 2;
  }
  return y;
}
long long ModInv(int a) { return ModExp(a, 1000000007 - 2); }
void Initialise() {
  fact[0] = 1 % 1000000007;
  for (int i = 1; i < 200001; ++i) fact[i] = (i * fact[i - 1]) % 1000000007;
  inv_fact[200001 - 1] = ModInv((int)fact[200001 - 1]);
  for (int i = 200001 - 2; i > -1; --i)
    inv_fact[i] = ((i + 1) * inv_fact[i + 1]) % 1000000007;
}
long long StirlingNumber2(int n, int k) {
  if (n < k) return 0ll;
  long long ans = 0;
  for (int j = 0, c = k % 2 ? -1 : 1; j <= k; ++j, c = -c) {
    long long ansx = (inv_fact[k - j] * inv_fact[j]) % 1000000007;
    ansx = (ansx * ModExp(j, n)) % 1000000007;
    ans = (ans + 1000000007 + c * ansx) % 1000000007;
  }
  return ans;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  long long s = 0;
  for (int i = 1; i <= n; ++i) {
    int j;
    scanf("%d", &j);
    s = (s + j) % 1000000007;
  }
  Initialise();
  long long c = (StirlingNumber2(n, k) + (n - 1) * StirlingNumber2(n - 1, k)) %
                1000000007;
  long long ans = (c * s) % 1000000007;
  printf("%lld\n", 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;
long long fact[200001], inv_fact[200001];
long long ModExp(int a, int n) {
  long long x = a, y = 1 % 1000000007;
  while (n) {
    if (n % 2) y = (y * x) % 1000000007;
    x = (x * x) % 1000000007;
    n /= 2;
  }
  return y;
}
long long ModInv(int a) { return ModExp(a, 1000000007 - 2); }
void Initialise() {
  fact[0] = 1 % 1000000007;
  for (int i = 1; i < 200001; ++i) fact[i] = (i * fact[i - 1]) % 1000000007;
  inv_fact[200001 - 1] = ModInv((int)fact[200001 - 1]);
  for (int i = 200001 - 2; i > -1; --i)
    inv_fact[i] = ((i + 1) * inv_fact[i + 1]) % 1000000007;
}
long long StirlingNumber2(int n, int k) {
  if (n < k) return 0ll;
  long long ans = 0;
  for (int j = 0, c = k % 2 ? -1 : 1; j <= k; ++j, c = -c) {
    long long ansx = (inv_fact[k - j] * inv_fact[j]) % 1000000007;
    ansx = (ansx * ModExp(j, n)) % 1000000007;
    ans = (ans + 1000000007 + c * ansx) % 1000000007;
  }
  return ans;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  long long s = 0;
  for (int i = 1; i <= n; ++i) {
    int j;
    scanf("%d", &j);
    s = (s + j) % 1000000007;
  }
  Initialise();
  long long c = (StirlingNumber2(n, k) + (n - 1) * StirlingNumber2(n - 1, k)) %
                1000000007;
  long long ans = (c * s) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}

```