#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7, MAXN = 2e5 + 3;
int times, K, N, jc[MAXN], inv[MAXN];
int poww(long long a, int b) {
  int times = 1;
  while (b) {
    if (b & 1) times = times * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return times;
}
int C(int b, int a) { return 1ll * jc[b] * inv[a] % MOD * inv[b - a] % MOD; }
int Stir(int b, int a) {
  int ans = 0;
  for (int i = 0; i < a; ++i)
    ans = (ans + MOD + (i & 1 ? -1ll : 1ll) * C(a, i) * poww(a - i, b) % MOD) %
          MOD;
  return 1ll * ans * inv[a] % MOD;
}
void calc() {
  jc[0] = 1;
  for (int i = 1; i <= N; ++i) jc[i] = 1ll * jc[i - 1] * i % MOD;
  inv[N] = poww(jc[N], MOD - 2);
  for (int i = N - 1; i >= 0; --i) inv[i] = inv[i + 1] * (i + 1ll) % MOD;
  times = (Stir(N, K) + (N - 1ll) * Stir(N - 1, K)) % MOD;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> N >> K;
  calc();
  int sum = 0;
  for (int i = 1; i <= N; ++i) {
    int a;
    cin >> a;
    sum = (sum + a) % MOD;
  }
  cout << 1ll * times * sum % MOD;
  return 0;
}

### Prompt

Your challenge is to write 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;
const int MOD = 1e9 + 7, MAXN = 2e5 + 3;
int times, K, N, jc[MAXN], inv[MAXN];
int poww(long long a, int b) {
  int times = 1;
  while (b) {
    if (b & 1) times = times * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return times;
}
int C(int b, int a) { return 1ll * jc[b] * inv[a] % MOD * inv[b - a] % MOD; }
int Stir(int b, int a) {
  int ans = 0;
  for (int i = 0; i < a; ++i)
    ans = (ans + MOD + (i & 1 ? -1ll : 1ll) * C(a, i) * poww(a - i, b) % MOD) %
          MOD;
  return 1ll * ans * inv[a] % MOD;
}
void calc() {
  jc[0] = 1;
  for (int i = 1; i <= N; ++i) jc[i] = 1ll * jc[i - 1] * i % MOD;
  inv[N] = poww(jc[N], MOD - 2);
  for (int i = N - 1; i >= 0; --i) inv[i] = inv[i + 1] * (i + 1ll) % MOD;
  times = (Stir(N, K) + (N - 1ll) * Stir(N - 1, K)) % MOD;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> N >> K;
  calc();
  int sum = 0;
  for (int i = 1; i <= N; ++i) {
    int a;
    cin >> a;
    sum = (sum + a) % MOD;
  }
  cout << 1ll * times * sum % MOD;
  return 0;
}

```