#include <bits/stdc++.h>
using namespace std;
const int N = 3e5, mod = 1000000007;
int ifac[N], pw[N], p[N], prime[N], n, k, cnt;
int qpower(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
void make(int n, int k) {
  pw[1] = ifac[1] = ifac[0] = 1;
  for (int i = 2; i <= n; i++) {
    ifac[i] = 1ll * ifac[mod % i] * (mod - mod / i) % mod;
    if (!p[i]) prime[++cnt] = i, pw[i] = qpower(i, k);
    for (int j = 1; j <= cnt && i * prime[j] <= n; j++) {
      int x = i * prime[j];
      p[x] = 1, pw[x] = 1ll * pw[i] * pw[prime[j]] % mod;
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 2; i <= n; i++) ifac[i] = 1ll * ifac[i - 1] * ifac[i] % mod;
}
int getin() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  return x;
}
int main() {
  scanf("%d%d", &n, &k);
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += getin();
  ans %= mod;
  int s = 0;
  make(k, n - 1);
  int C = 1;
  for (int i = 0; i <= k; i++)
    s = (s + (((k - i) & 1) ? -1 : 1) * 1ll * ifac[i] * ifac[k - i] % mod *
                 pw[i] % mod * (i + n - 1)) %
        mod;
  cout << ans * (s + mod) % mod << endl;
}

### Prompt

Generate 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 = 3e5, mod = 1000000007;
int ifac[N], pw[N], p[N], prime[N], n, k, cnt;
int qpower(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
void make(int n, int k) {
  pw[1] = ifac[1] = ifac[0] = 1;
  for (int i = 2; i <= n; i++) {
    ifac[i] = 1ll * ifac[mod % i] * (mod - mod / i) % mod;
    if (!p[i]) prime[++cnt] = i, pw[i] = qpower(i, k);
    for (int j = 1; j <= cnt && i * prime[j] <= n; j++) {
      int x = i * prime[j];
      p[x] = 1, pw[x] = 1ll * pw[i] * pw[prime[j]] % mod;
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 2; i <= n; i++) ifac[i] = 1ll * ifac[i - 1] * ifac[i] % mod;
}
int getin() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  return x;
}
int main() {
  scanf("%d%d", &n, &k);
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += getin();
  ans %= mod;
  int s = 0;
  make(k, n - 1);
  int C = 1;
  for (int i = 0; i <= k; i++)
    s = (s + (((k - i) & 1) ? -1 : 1) * 1ll * ifac[i] * ifac[k - i] % mod *
                 pw[i] % mod * (i + n - 1)) %
        mod;
  cout << ans * (s + mod) % mod << endl;
}

```