#include <bits/stdc++.h>
using namespace std;
long long n, k, nums[200200];
long long fat[200200], ifat[200200];
long long fpow(long long base, long long exp) {
  if (exp < 1) return 1;
  long long ret = fpow(base, exp / 2);
  ret *= ret;
  ret %= 1000000007;
  if (exp & 1) ret *= base;
  return ret % 1000000007;
}
long long comb(long long a, long long b) {
  return (((fat[a] * ifat[b]) % 1000000007) * ifat[a - b]) % 1000000007;
}
int main() {
  scanf("%lld %lld", &n, &k);
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &nums[i]);
    sum += nums[i];
  }
  sum %= 1000000007;
  fat[0] = 1;
  for (int i = 1; i < 200200; i++) {
    fat[i] = fat[i - 1] * i;
    fat[i] %= 1000000007;
  }
  for (int i = 0; i < 200200; i++) ifat[i] = fpow(fat[i], 1000000007 - 2);
  long long val1 = 0, val2 = 0;
  for (int i = 0; i <= k; i++) {
    long long aux = 1;
    if ((i + k) & 1) aux *= -1;
    aux *= comb(k, i);
    aux *= fpow(i, n);
    aux %= 1000000007;
    val1 += aux;
    val1 %= 1000000007;
    aux = 1;
    if ((i + k) & 1) aux *= -1;
    aux *= comb(k, i);
    aux *= fpow(i, n - 1);
    aux %= 1000000007;
    val2 += aux;
    val2 %= 1000000007;
  }
  val1 *= ifat[k];
  val1 %= 1000000007;
  val2 *= ifat[k];
  val2 %= 1000000007;
  long long ans = val1 + val2 * (n - 1);
  ans %= 1000000007;
  ans += 1000000007;
  ans %= 1000000007;
  printf("%lld\n", (ans * sum) % 1000000007);
  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 n, k, nums[200200];
long long fat[200200], ifat[200200];
long long fpow(long long base, long long exp) {
  if (exp < 1) return 1;
  long long ret = fpow(base, exp / 2);
  ret *= ret;
  ret %= 1000000007;
  if (exp & 1) ret *= base;
  return ret % 1000000007;
}
long long comb(long long a, long long b) {
  return (((fat[a] * ifat[b]) % 1000000007) * ifat[a - b]) % 1000000007;
}
int main() {
  scanf("%lld %lld", &n, &k);
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &nums[i]);
    sum += nums[i];
  }
  sum %= 1000000007;
  fat[0] = 1;
  for (int i = 1; i < 200200; i++) {
    fat[i] = fat[i - 1] * i;
    fat[i] %= 1000000007;
  }
  for (int i = 0; i < 200200; i++) ifat[i] = fpow(fat[i], 1000000007 - 2);
  long long val1 = 0, val2 = 0;
  for (int i = 0; i <= k; i++) {
    long long aux = 1;
    if ((i + k) & 1) aux *= -1;
    aux *= comb(k, i);
    aux *= fpow(i, n);
    aux %= 1000000007;
    val1 += aux;
    val1 %= 1000000007;
    aux = 1;
    if ((i + k) & 1) aux *= -1;
    aux *= comb(k, i);
    aux *= fpow(i, n - 1);
    aux %= 1000000007;
    val2 += aux;
    val2 %= 1000000007;
  }
  val1 *= ifat[k];
  val1 %= 1000000007;
  val2 *= ifat[k];
  val2 %= 1000000007;
  long long ans = val1 + val2 * (n - 1);
  ans %= 1000000007;
  ans += 1000000007;
  ans %= 1000000007;
  printf("%lld\n", (ans * sum) % 1000000007);
  return 0;
}

```