#include <bits/stdc++.h>
using namespace std;
const int M = 200006;
int f[M], inv[M];
int expo(int base, int exponent, int mod) {
  int ans = 1;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base) % mod;
    base = (1LL * base * base) % mod;
    exponent >>= 1;
  }
  return ans % mod;
}
void compute() {
  f[0] = 1;
  for (int i = 1; i < M; i++) {
    f[i] = (1LL * i * f[i - 1]) % 1000000007;
  }
  inv[M - 1] = expo(f[M - 1], 1000000007 - 2, 1000000007);
  for (int i = M - 2; i >= 0; i--) {
    inv[i] = (1LL * (i + 1) * inv[i + 1]) % 1000000007;
  }
}
int C(int n, int r) {
  return (1LL * ((1LL * f[n] * inv[r]) % 1000000007) * inv[n - r]) % 1000000007;
}
int S(int n, int k) {
  int ans = 0;
  for (int i = 0; i < k + 1; i++) {
    if (i % 2) {
      ans = (ans + 1000000007 -
             1LL * C(k, i) * expo(k - i, n, 1000000007) % 1000000007) %
            1000000007;
    } else {
      ans = (ans + 1LL * C(k, i) * expo(k - i, n, 1000000007)) % 1000000007;
    }
  }
  return (1LL * ans * inv[k]) % 1000000007;
}
void solve() {
  compute();
  int k, n, s = 0;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int first;
    cin >> first;
    s += first;
    s %= 1000000007;
  }
  cout << 1LL * s * (S(n, k) + 1LL * (n - 1) * S(n - 1, k) % 1000000007) %
              1000000007
       << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  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;
const int M = 200006;
int f[M], inv[M];
int expo(int base, int exponent, int mod) {
  int ans = 1;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base) % mod;
    base = (1LL * base * base) % mod;
    exponent >>= 1;
  }
  return ans % mod;
}
void compute() {
  f[0] = 1;
  for (int i = 1; i < M; i++) {
    f[i] = (1LL * i * f[i - 1]) % 1000000007;
  }
  inv[M - 1] = expo(f[M - 1], 1000000007 - 2, 1000000007);
  for (int i = M - 2; i >= 0; i--) {
    inv[i] = (1LL * (i + 1) * inv[i + 1]) % 1000000007;
  }
}
int C(int n, int r) {
  return (1LL * ((1LL * f[n] * inv[r]) % 1000000007) * inv[n - r]) % 1000000007;
}
int S(int n, int k) {
  int ans = 0;
  for (int i = 0; i < k + 1; i++) {
    if (i % 2) {
      ans = (ans + 1000000007 -
             1LL * C(k, i) * expo(k - i, n, 1000000007) % 1000000007) %
            1000000007;
    } else {
      ans = (ans + 1LL * C(k, i) * expo(k - i, n, 1000000007)) % 1000000007;
    }
  }
  return (1LL * ans * inv[k]) % 1000000007;
}
void solve() {
  compute();
  int k, n, s = 0;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int first;
    cin >> first;
    s += first;
    s %= 1000000007;
  }
  cout << 1LL * s * (S(n, k) + 1LL * (n - 1) * S(n - 1, k) % 1000000007) %
              1000000007
       << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```