#include <bits/stdc++.h>
using namespace std;
const long long int modulo = 1e9 + 7;
void anyade(long long int &a, long long int b) { a = (a + b) % modulo; }
const int limite = 1000005;
int n, k;
long long int sw;
long long int fact[limite];
long long int factinv[limite];
long long int eleva(long long int b, long long int e) {
  if (e == 0) return 1;
  if (e % 2 == 0) return eleva(b * b % modulo, e / 2);
  return b * eleva(b * b % modulo, e / 2) % modulo;
}
long long int inverso(long long int x) { return eleva(x, modulo - 2); }
long long int combi(long long int n, long long int k) {
  if (k < 0 or n < k) return 0;
  return fact[n] * factinv[k] % modulo * factinv[n - k] % modulo;
}
long long int part(long long int n, long long int k) {
  if (n < 0) return 0;
  if (k < 0) return 0;
  if (k == 0) return n == k;
  if (n < k) return 0;
  if (n == k) return 1;
  long long int sol = 0;
  long long int signo = 1;
  for (int i = 0; i < k; i++, signo *= -1)
    anyade(sol, signo * combi(k, i) * eleva(k - i, n));
  sol = sol * factinv[k] % modulo;
  return sol;
}
long long int sol;
int main() {
  fact[0] = factinv[0] = 1;
  for (int i = 1; i < limite; i++) {
    fact[i] = i * fact[i - 1] % modulo;
    factinv[i] = inverso(i) * factinv[i - 1] % modulo;
  }
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int w;
    scanf("%d", &w);
    anyade(sw, w);
  }
  anyade(sol, part(n, k));
  anyade(sol, (n - 1) * part(n - 1, k));
  sol = sol * sw % modulo;
  sol = ((sol % modulo) + modulo) % modulo;
  printf("%d\n", int(sol));
}

### Prompt

Please create a solution in Cpp 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 long long int modulo = 1e9 + 7;
void anyade(long long int &a, long long int b) { a = (a + b) % modulo; }
const int limite = 1000005;
int n, k;
long long int sw;
long long int fact[limite];
long long int factinv[limite];
long long int eleva(long long int b, long long int e) {
  if (e == 0) return 1;
  if (e % 2 == 0) return eleva(b * b % modulo, e / 2);
  return b * eleva(b * b % modulo, e / 2) % modulo;
}
long long int inverso(long long int x) { return eleva(x, modulo - 2); }
long long int combi(long long int n, long long int k) {
  if (k < 0 or n < k) return 0;
  return fact[n] * factinv[k] % modulo * factinv[n - k] % modulo;
}
long long int part(long long int n, long long int k) {
  if (n < 0) return 0;
  if (k < 0) return 0;
  if (k == 0) return n == k;
  if (n < k) return 0;
  if (n == k) return 1;
  long long int sol = 0;
  long long int signo = 1;
  for (int i = 0; i < k; i++, signo *= -1)
    anyade(sol, signo * combi(k, i) * eleva(k - i, n));
  sol = sol * factinv[k] % modulo;
  return sol;
}
long long int sol;
int main() {
  fact[0] = factinv[0] = 1;
  for (int i = 1; i < limite; i++) {
    fact[i] = i * fact[i - 1] % modulo;
    factinv[i] = inverso(i) * factinv[i - 1] % modulo;
  }
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int w;
    scanf("%d", &w);
    anyade(sw, w);
  }
  anyade(sol, part(n, k));
  anyade(sol, (n - 1) * part(n - 1, k));
  sol = sol * sw % modulo;
  sol = ((sol % modulo) + modulo) % modulo;
  printf("%d\n", int(sol));
}

```