#include <bits/stdc++.h>
using namespace std;
long long inv[500005], n, k, x, sum;
inline long long ksm(long long x, long long y) {
  long long ans1 = 1;
  while (y) {
    if (y & 1) ans1 = 1ll * ans1 * x % 1000000007;
    x = 1ll * x * x % 1000000007;
    y >>= 1;
  }
  return ans1;
}
inline long long S(long long n, long long m) {
  long long ans = 0;
  for (long long i = 0; i <= m; i++) {
    long long t =
        1ll * inv[i] * inv[m - i] % 1000000007 * ksm(m - i, n) % 1000000007;
    if (i & 1)
      ans = 1ll * (ans - t + 1000000007) % 1000000007;
    else
      ans = 1ll * (ans + t) % 1000000007;
  }
  return ans;
}
int main() {
  scanf("%lld%lld", &n, &k);
  inv[0] = inv[1] = 1;
  for (long long i = 2; i <= k; i++)
    inv[i] = -1ll * 1000000007 / i * inv[1000000007 % i] % 1000000007;
  for (long long i = 1; i <= k; i++)
    inv[i] = 1ll * inv[i - 1] * inv[i] % 1000000007;
  for (long long i = 1; i <= n; i++)
    scanf("%lld", &x), sum = 1ll * (sum + x) % 1000000007;
  long long ans =
      1ll * (S(n - 1, k - 1) + (1ll * (n + k - 1) * S(n - 1, k) % 1000000007)) %
      1000000007;
  ans = 1ll * ans * sum % 1000000007;
  printf("%lld\n", (ans % 1000000007 + 1000000007) % 1000000007);
  return 0;
}

### Prompt

Please formulate 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;
long long inv[500005], n, k, x, sum;
inline long long ksm(long long x, long long y) {
  long long ans1 = 1;
  while (y) {
    if (y & 1) ans1 = 1ll * ans1 * x % 1000000007;
    x = 1ll * x * x % 1000000007;
    y >>= 1;
  }
  return ans1;
}
inline long long S(long long n, long long m) {
  long long ans = 0;
  for (long long i = 0; i <= m; i++) {
    long long t =
        1ll * inv[i] * inv[m - i] % 1000000007 * ksm(m - i, n) % 1000000007;
    if (i & 1)
      ans = 1ll * (ans - t + 1000000007) % 1000000007;
    else
      ans = 1ll * (ans + t) % 1000000007;
  }
  return ans;
}
int main() {
  scanf("%lld%lld", &n, &k);
  inv[0] = inv[1] = 1;
  for (long long i = 2; i <= k; i++)
    inv[i] = -1ll * 1000000007 / i * inv[1000000007 % i] % 1000000007;
  for (long long i = 1; i <= k; i++)
    inv[i] = 1ll * inv[i - 1] * inv[i] % 1000000007;
  for (long long i = 1; i <= n; i++)
    scanf("%lld", &x), sum = 1ll * (sum + x) % 1000000007;
  long long ans =
      1ll * (S(n - 1, k - 1) + (1ll * (n + k - 1) * S(n - 1, k) % 1000000007)) %
      1000000007;
  ans = 1ll * ans * sum % 1000000007;
  printf("%lld\n", (ans % 1000000007 + 1000000007) % 1000000007);
  return 0;
}

```