#include <bits/stdc++.h>
using namespace std;
constexpr int inf32 = 0x3f3f3f3f;
constexpr long long inf64 = 0x3f3f3f3f3f3f3f3f;
constexpr int mod = 1000000007;
struct mint {
  int first;
  mint(int first = 0) : first(first) {}
  friend mint operator+(const mint a, const mint &b) {
    int first = a.first + b.first;
    return mint(first >= mod ? first - mod : first);
  }
  friend mint operator-(const mint a, const mint &b) {
    int first = a.first - b.first;
    return mint(first < 0 ? first + mod : first);
  }
  friend mint operator*(const mint a, const mint &b) {
    return mint(1ll * a.first * b.first % mod);
  }
};
int mypow(int a, int first = mod - 2, int res = 1) {
  for (; first; first >>= 1, a = (mint(a) * a).first)
    if (first & 1) res = (mint(res) * a).first;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  int sm = 0;
  for (int i = 0; i < (n); ++i) {
    int first;
    cin >> first;
    sm = (mint(sm) + first).first;
  }
  if (n == 1) return !(cout << sm << '\n');
  vector<int> fac(k), ifac(k);
  fac[0] = 1;
  for (int i = 0; i < (k); ++i)
    if (i) fac[i] = (mint(fac[i - 1]) * i).first;
  ifac[k - 1] = mypow(fac[k - 1]);
  for (int i = (k)-1; i >= 0; --i)
    if (i) ifac[i - 1] = (mint(ifac[i]) * i).first;
  int c = 0;
  for (int i = 0; i < (k); ++i) {
    int s = ((mint(n) - 1 + k - i) * mypow(k - i, n - 2) * ifac[i] *
             ifac[k - 1 - i])
                .first;
    if (i & 1)
      c = (mint(c) - s).first;
    else
      c = (mint(c) + s).first;
  }
  c = (mint(c) * sm).first;
  cout << c << '\n';
  return 0;
}

### Prompt

Your task is to create 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;
constexpr int inf32 = 0x3f3f3f3f;
constexpr long long inf64 = 0x3f3f3f3f3f3f3f3f;
constexpr int mod = 1000000007;
struct mint {
  int first;
  mint(int first = 0) : first(first) {}
  friend mint operator+(const mint a, const mint &b) {
    int first = a.first + b.first;
    return mint(first >= mod ? first - mod : first);
  }
  friend mint operator-(const mint a, const mint &b) {
    int first = a.first - b.first;
    return mint(first < 0 ? first + mod : first);
  }
  friend mint operator*(const mint a, const mint &b) {
    return mint(1ll * a.first * b.first % mod);
  }
};
int mypow(int a, int first = mod - 2, int res = 1) {
  for (; first; first >>= 1, a = (mint(a) * a).first)
    if (first & 1) res = (mint(res) * a).first;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  int sm = 0;
  for (int i = 0; i < (n); ++i) {
    int first;
    cin >> first;
    sm = (mint(sm) + first).first;
  }
  if (n == 1) return !(cout << sm << '\n');
  vector<int> fac(k), ifac(k);
  fac[0] = 1;
  for (int i = 0; i < (k); ++i)
    if (i) fac[i] = (mint(fac[i - 1]) * i).first;
  ifac[k - 1] = mypow(fac[k - 1]);
  for (int i = (k)-1; i >= 0; --i)
    if (i) ifac[i - 1] = (mint(ifac[i]) * i).first;
  int c = 0;
  for (int i = 0; i < (k); ++i) {
    int s = ((mint(n) - 1 + k - i) * mypow(k - i, n - 2) * ifac[i] *
             ifac[k - 1 - i])
                .first;
    if (i & 1)
      c = (mint(c) - s).first;
    else
      c = (mint(c) + s).first;
  }
  c = (mint(c) * sm).first;
  cout << c << '\n';
  return 0;
}

```