#include <bits/stdc++.h>
inline void read(int &x) {
  x = 0;
  register char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
}
int n, w;
struct edge {
  int nxt, v;
} e[200005 << 1];
int head[200005], tot, k;
inline void addedge(int u, int v) {
  e[++tot] = (edge){head[u], v};
  head[u] = tot;
}
long long invfac[200005], fac[200005], ans, sum;
inline long long pow(long long x, int y) {
  long long ret = 1;
  if (y < 0) return 0;
  while (y) {
    if (y & 1) ret = ret * x % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return ret;
}
inline long long C(int n, int m) {
  return fac[n] * invfac[m] % 1000000007 * invfac[n - m] % 1000000007;
}
inline long long second(int n, int m) {
  long long ret = 0;
  for (int k = (0); k <= (m); ++k)
    (ret += (k & 1 ? -1 : 1) * invfac[k] * invfac[m - k] % 1000000007 *
            pow(m - k, n)) %= 1000000007;
  return ret;
}
int main() {
  read(n);
  read(k);
  for (int i = (1); i <= (n); ++i) {
    read(w);
    sum = (sum + w) % 1000000007;
  }
  fac[0] = invfac[0] = 1;
  for (int i = (1); i <= (n); ++i) fac[i] = fac[i - 1] * i % 1000000007;
  invfac[n] = pow(fac[n], 1000000007 - 2);
  for (int i = (n - 1); i >= (1); --i)
    invfac[i] = invfac[i + 1] * (i + 1) % 1000000007;
  printf("%lld\n", ((second(n - 1, k - 1) + second(n - 1, k) * (n + k - 1)) %
                        1000000007 * sum % 1000000007 +
                    1000000007) %
                       1000000007);
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
inline void read(int &x) {
  x = 0;
  register char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
}
int n, w;
struct edge {
  int nxt, v;
} e[200005 << 1];
int head[200005], tot, k;
inline void addedge(int u, int v) {
  e[++tot] = (edge){head[u], v};
  head[u] = tot;
}
long long invfac[200005], fac[200005], ans, sum;
inline long long pow(long long x, int y) {
  long long ret = 1;
  if (y < 0) return 0;
  while (y) {
    if (y & 1) ret = ret * x % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return ret;
}
inline long long C(int n, int m) {
  return fac[n] * invfac[m] % 1000000007 * invfac[n - m] % 1000000007;
}
inline long long second(int n, int m) {
  long long ret = 0;
  for (int k = (0); k <= (m); ++k)
    (ret += (k & 1 ? -1 : 1) * invfac[k] * invfac[m - k] % 1000000007 *
            pow(m - k, n)) %= 1000000007;
  return ret;
}
int main() {
  read(n);
  read(k);
  for (int i = (1); i <= (n); ++i) {
    read(w);
    sum = (sum + w) % 1000000007;
  }
  fac[0] = invfac[0] = 1;
  for (int i = (1); i <= (n); ++i) fac[i] = fac[i - 1] * i % 1000000007;
  invfac[n] = pow(fac[n], 1000000007 - 2);
  for (int i = (n - 1); i >= (1); --i)
    invfac[i] = invfac[i + 1] * (i + 1) % 1000000007;
  printf("%lld\n", ((second(n - 1, k - 1) + second(n - 1, k) * (n + k - 1)) %
                        1000000007 * sum % 1000000007 +
                    1000000007) %
                       1000000007);
  return 0;
}

```