#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = false;
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  if (flag) x = -x;
}
long long n, k, s, ans;
long long fac[200010], ifac[200010];
long long qp(long long x, long long y) {
  long long v = 1;
  while (y) {
    if (y & 1) v = v * x % 1000000007;
    x = x * x % 1000000007, y >>= 1;
  }
  return v;
}
int main() {
  read(n), read(k), fac[0] = ifac[0] = 1;
  for (int i = 1, x; i <= n; ++i) read(x), s = (s + x) % 1000000007;
  if (n == 1) {
    printf("%lld\n", s);
    return 0;
  }
  for (int i = 1; i <= k; ++i) fac[i] = fac[i - 1] * i % 1000000007;
  ifac[k] = qp(fac[k], 1000000007 - 2);
  for (int i = k - 1; i; --i) ifac[i] = ifac[i + 1] * (i + 1) % 1000000007;
  for (int i = 0; i <= k - 1; ++i)
    ans = (ans +
           ((k - 1 - i) & 1 ? -1 : 1) * ifac[i] * ifac[k - 1 - i] % 1000000007 *
               qp(i + 1, n - 2) % 1000000007 * (n + i) % 1000000007 +
           1000000007) %
          1000000007;
  printf("%lld", ans * s % 1000000007);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = false;
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  if (flag) x = -x;
}
long long n, k, s, ans;
long long fac[200010], ifac[200010];
long long qp(long long x, long long y) {
  long long v = 1;
  while (y) {
    if (y & 1) v = v * x % 1000000007;
    x = x * x % 1000000007, y >>= 1;
  }
  return v;
}
int main() {
  read(n), read(k), fac[0] = ifac[0] = 1;
  for (int i = 1, x; i <= n; ++i) read(x), s = (s + x) % 1000000007;
  if (n == 1) {
    printf("%lld\n", s);
    return 0;
  }
  for (int i = 1; i <= k; ++i) fac[i] = fac[i - 1] * i % 1000000007;
  ifac[k] = qp(fac[k], 1000000007 - 2);
  for (int i = k - 1; i; --i) ifac[i] = ifac[i + 1] * (i + 1) % 1000000007;
  for (int i = 0; i <= k - 1; ++i)
    ans = (ans +
           ((k - 1 - i) & 1 ? -1 : 1) * ifac[i] * ifac[k - 1 - i] % 1000000007 *
               qp(i + 1, n - 2) % 1000000007 * (n + i) % 1000000007 +
           1000000007) %
          1000000007;
  printf("%lld", ans * s % 1000000007);
  return 0;
}

```