#include <bits/stdc++.h>
using namespace std;
long long int fact[1000005];
long long int ifact[1000005];
long long int bexpo(long long int x, long long int n, long long int m) {
  long long int r = 1;
  x = x % m;
  while (n > 0) {
    if (n % 2) r = (r * x) % m;
    x = (x * x) % m;
    n = n / 2;
  }
  return r;
}
long long int ncr(long long int n, long long int r) {
  if (r > n) return 0;
  long long int ans = fact[n];
  ans = (ans * ifact[r]) % 1000000007;
  ans = (ans * ifact[n - r]) % 1000000007;
  return ans;
}
void pre() {
  fact[0] = 1;
  ifact[0] = 1;
  for (long long int i = 1; i < 1000005; i++)
    fact[i] = (fact[i - 1] * i) % 1000000007;
  ifact[1000005 - 1] = bexpo(fact[1000005 - 1], 1000000007 - 2, 1000000007);
  for (long long int i = 1000005 - 2; i > 0; i--)
    ifact[i] = (ifact[i + 1] * (i + 1)) % 1000000007;
}
long long int fn(long long int n, long long int k) {
  long long int ans = 0;
  for (long long int i = 0; i <= k; i++) {
    long long int temp = ncr(k, i);
    temp = (temp * bexpo(i, n, 1000000007)) % 1000000007;
    if ((k + i) % 2) {
      temp = temp * -1;
      temp %= 1000000007;
      temp = (temp + 1000000007) % 1000000007;
    }
    ans = (ans + temp) % 1000000007;
  }
  ans = (ans * ifact[k]) % 1000000007;
  return ans;
}
int main() {
  pre();
  long long int n, k;
  scanf("%lld%lld", &n, &k);
  long long int sum = 0;
  for (long long int i = 1; i <= n; i++) {
    long long int d;
    scanf("%lld", &d);
    sum = (sum + d) % 1000000007;
  }
  long long int temp = fn(n, k);
  temp = (temp + (n - 1) * (fn(n - 1, k)) % 1000000007) % 1000000007;
  sum = (sum * temp) % 1000000007;
  printf("%lld\n", sum);
  return 0;
}

### Prompt

Please formulate 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;
long long int fact[1000005];
long long int ifact[1000005];
long long int bexpo(long long int x, long long int n, long long int m) {
  long long int r = 1;
  x = x % m;
  while (n > 0) {
    if (n % 2) r = (r * x) % m;
    x = (x * x) % m;
    n = n / 2;
  }
  return r;
}
long long int ncr(long long int n, long long int r) {
  if (r > n) return 0;
  long long int ans = fact[n];
  ans = (ans * ifact[r]) % 1000000007;
  ans = (ans * ifact[n - r]) % 1000000007;
  return ans;
}
void pre() {
  fact[0] = 1;
  ifact[0] = 1;
  for (long long int i = 1; i < 1000005; i++)
    fact[i] = (fact[i - 1] * i) % 1000000007;
  ifact[1000005 - 1] = bexpo(fact[1000005 - 1], 1000000007 - 2, 1000000007);
  for (long long int i = 1000005 - 2; i > 0; i--)
    ifact[i] = (ifact[i + 1] * (i + 1)) % 1000000007;
}
long long int fn(long long int n, long long int k) {
  long long int ans = 0;
  for (long long int i = 0; i <= k; i++) {
    long long int temp = ncr(k, i);
    temp = (temp * bexpo(i, n, 1000000007)) % 1000000007;
    if ((k + i) % 2) {
      temp = temp * -1;
      temp %= 1000000007;
      temp = (temp + 1000000007) % 1000000007;
    }
    ans = (ans + temp) % 1000000007;
  }
  ans = (ans * ifact[k]) % 1000000007;
  return ans;
}
int main() {
  pre();
  long long int n, k;
  scanf("%lld%lld", &n, &k);
  long long int sum = 0;
  for (long long int i = 1; i <= n; i++) {
    long long int d;
    scanf("%lld", &d);
    sum = (sum + d) % 1000000007;
  }
  long long int temp = fn(n, k);
  temp = (temp + (n - 1) * (fn(n - 1, k)) % 1000000007) % 1000000007;
  sum = (sum * temp) % 1000000007;
  printf("%lld\n", sum);
  return 0;
}

```