#include <bits/stdc++.h>
using namespace std;
long long gi() {
  long long ret = 0;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  char c = ch == '-' ? getchar() : ch;
  while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
  return ch == '-' ? -ret : ret;
}
const int mo = 1e9 + 7;
const int N = 200010;
int n, k, a[N];
long long fac[N], ifac[N];
long long ksm(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mo;
    a = a * a % mo;
    b >>= 1;
  }
  return ret;
}
long long C(int n, int m) {
  if (n < m) return 0;
  return fac[n] * ifac[m] % mo * ifac[n - m] % mo;
}
void calcS(long long &s, int k) {
  for (int i = 0; i <= k; i += 1)
    (s += ((i & 1) ? -1 : 1) * C(k, i) % mo * ksm(k - i, n - 1)) %= mo;
  s = s * ifac[k] % mo;
}
int main() {
  n = gi();
  k = gi();
  for (int i = 1; i <= n; i += 1) a[i] = gi();
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= n; i += 1) fac[i] = fac[i - 1] * i % mo;
  ifac[n] = ksm(fac[n], mo - 2);
  for (int i = n; i >= 1; i -= 1) ifac[i - 1] = ifac[i] * i % mo;
  long long s1 = 0, s2 = 0;
  calcS(s1, k - 1);
  calcS(s2, k);
  long long ans = 0;
  for (int i = 1; i <= n; i += 1)
    (ans += s1 * a[i] + s2 * (n - 1 + k) % mo * a[i]) %= mo;
  cout << (ans + mo) % mo;
  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;
long long gi() {
  long long ret = 0;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  char c = ch == '-' ? getchar() : ch;
  while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
  return ch == '-' ? -ret : ret;
}
const int mo = 1e9 + 7;
const int N = 200010;
int n, k, a[N];
long long fac[N], ifac[N];
long long ksm(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mo;
    a = a * a % mo;
    b >>= 1;
  }
  return ret;
}
long long C(int n, int m) {
  if (n < m) return 0;
  return fac[n] * ifac[m] % mo * ifac[n - m] % mo;
}
void calcS(long long &s, int k) {
  for (int i = 0; i <= k; i += 1)
    (s += ((i & 1) ? -1 : 1) * C(k, i) % mo * ksm(k - i, n - 1)) %= mo;
  s = s * ifac[k] % mo;
}
int main() {
  n = gi();
  k = gi();
  for (int i = 1; i <= n; i += 1) a[i] = gi();
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= n; i += 1) fac[i] = fac[i - 1] * i % mo;
  ifac[n] = ksm(fac[n], mo - 2);
  for (int i = n; i >= 1; i -= 1) ifac[i - 1] = ifac[i] * i % mo;
  long long s1 = 0, s2 = 0;
  calcS(s1, k - 1);
  calcS(s2, k);
  long long ans = 0;
  for (int i = 1; i <= n; i += 1)
    (ans += s1 * a[i] + s2 * (n - 1 + k) % mo * a[i]) %= mo;
  cout << (ans + mo) % mo;
  return 0;
}

```