#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
long long fac[maxn];
long long ifac[maxn];
void initial(const int n = 2e5) {
  fac[0] = 1;
  fac[1] = 1;
  for (int i = 2; i <= n; i++) fac[i] = 1ll * i * fac[i - 1] % mod;
  ifac[0] = 1;
  ifac[1] = 1;
  for (int i = 2; i <= n; i++) ifac[i] = (mod - mod / i) * ifac[mod % i] % mod;
  for (int i = 2; i <= n; i++) ifac[i] = ifac[i] * ifac[i - 1] % mod;
}
long long C(int n, int m) { return fac[n] * ifac[n - m] % mod * ifac[m] % mod; }
long long _pow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
long long s(int n, int k) {
  long long res = 0;
  for (int j = 0; j <= k; j++) {
    if ((j + k) & 1)
      res = (res - C(k, j) * _pow(j, n) % mod + mod) % mod;
    else
      res = (res + C(k, j) * _pow(j, n) % mod) % mod;
  }
  res = res * ifac[k] % mod;
  return res;
}
int main(void) {
  initial();
  int n, k;
  scanf("%d%d", &n, &k);
  long long res = s(n, k) + 1ll * (n - 1) * s(n - 1, k) % mod;
  res %= mod;
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    sum = (sum + x) % mod;
  }
  res = 1ll * res * sum % mod;
  printf("%lld\n", res);
  return 0;
}

### 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 maxn = 2e5 + 5;
const int mod = 1e9 + 7;
long long fac[maxn];
long long ifac[maxn];
void initial(const int n = 2e5) {
  fac[0] = 1;
  fac[1] = 1;
  for (int i = 2; i <= n; i++) fac[i] = 1ll * i * fac[i - 1] % mod;
  ifac[0] = 1;
  ifac[1] = 1;
  for (int i = 2; i <= n; i++) ifac[i] = (mod - mod / i) * ifac[mod % i] % mod;
  for (int i = 2; i <= n; i++) ifac[i] = ifac[i] * ifac[i - 1] % mod;
}
long long C(int n, int m) { return fac[n] * ifac[n - m] % mod * ifac[m] % mod; }
long long _pow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
long long s(int n, int k) {
  long long res = 0;
  for (int j = 0; j <= k; j++) {
    if ((j + k) & 1)
      res = (res - C(k, j) * _pow(j, n) % mod + mod) % mod;
    else
      res = (res + C(k, j) * _pow(j, n) % mod) % mod;
  }
  res = res * ifac[k] % mod;
  return res;
}
int main(void) {
  initial();
  int n, k;
  scanf("%d%d", &n, &k);
  long long res = s(n, k) + 1ll * (n - 1) * s(n - 1, k) % mod;
  res %= mod;
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    sum = (sum + x) % mod;
  }
  res = 1ll * res * sum % mod;
  printf("%lld\n", res);
  return 0;
}

```