#include <bits/stdc++.h>
using namespace std;
int fact[200010], inv[200010];
int sum(int a, int b) {
  return (a + b >= 1000000007 ? a + b - 1000000007 : a + b);
}
int dif(int a, int b) { return (a < b ? a - b + 1000000007 : a - b); }
int prod(int a, int b) { return (1LL * a * b) % 1000000007; }
int power(int base, int exp) {
  int res;
  for (res = 1; exp; exp >>= 1) {
    if (exp & 1) res = prod(res, base);
    base = prod(base, base);
  }
  return res;
}
int comb(int n, int k) { return prod(prod(fact[n], inv[k]), inv[n - k]); }
int stirling(int n, int k) {
  if (n < k) return 0;
  int res, term, sign, j;
  for (res = 0, sign = ((k & 1) ? -1 : 1), j = 0; j <= k; ++j, sign = -sign) {
    term = prod(comb(k, j), power(j, n));
    if (sign > 0)
      res = sum(res, term);
    else
      res = dif(res, term);
  }
  return prod(res, inv[k]);
}
int main() {
  int i, k, n, s, w;
  for (fact[0] = 1, i = 1; i < 200010; ++i) fact[i] = prod(fact[i - 1], i);
  for (inv[200010 - 1] = power(fact[200010 - 1], 1000000007 - 2),
                    i = 200010 - 2;
       i >= 0; --i)
    inv[i] = prod(i + 1, inv[i + 1]);
  cin >> n >> k;
  for (s = 0, i = 1; i <= n; ++i) {
    cin >> w;
    s = sum(s, w);
  }
  cout << prod(s, sum(stirling(n, k), prod(n - 1, stirling(n - 1, k)))) << '\n';
  return 0;
}

### 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;
int fact[200010], inv[200010];
int sum(int a, int b) {
  return (a + b >= 1000000007 ? a + b - 1000000007 : a + b);
}
int dif(int a, int b) { return (a < b ? a - b + 1000000007 : a - b); }
int prod(int a, int b) { return (1LL * a * b) % 1000000007; }
int power(int base, int exp) {
  int res;
  for (res = 1; exp; exp >>= 1) {
    if (exp & 1) res = prod(res, base);
    base = prod(base, base);
  }
  return res;
}
int comb(int n, int k) { return prod(prod(fact[n], inv[k]), inv[n - k]); }
int stirling(int n, int k) {
  if (n < k) return 0;
  int res, term, sign, j;
  for (res = 0, sign = ((k & 1) ? -1 : 1), j = 0; j <= k; ++j, sign = -sign) {
    term = prod(comb(k, j), power(j, n));
    if (sign > 0)
      res = sum(res, term);
    else
      res = dif(res, term);
  }
  return prod(res, inv[k]);
}
int main() {
  int i, k, n, s, w;
  for (fact[0] = 1, i = 1; i < 200010; ++i) fact[i] = prod(fact[i - 1], i);
  for (inv[200010 - 1] = power(fact[200010 - 1], 1000000007 - 2),
                    i = 200010 - 2;
       i >= 0; --i)
    inv[i] = prod(i + 1, inv[i + 1]);
  cin >> n >> k;
  for (s = 0, i = 1; i <= n; ++i) {
    cin >> w;
    s = sum(s, w);
  }
  cout << prod(s, sum(stirling(n, k), prod(n - 1, stirling(n - 1, k)))) << '\n';
  return 0;
}

```