#include <bits/stdc++.h>
using namespace std;
long long int pw(long long int a, long long int b, long long int mod) {
  if (!b) return 1;
  if (b & 1) return a * pw(a * a % mod, b / 2, mod) % mod;
  return pw(a * a % mod, b / 2, mod) % mod;
}
const long long int N = 2e5 + 10;
const long long int MOD = 1e9 + 7;
const long long int INF = 1e18;
long long int n, k, A[N], sum, g, fact[N], inv[N], ans;
inline long long int C(long long int n, long long int r) {
  if (r < 0 || n < r) return 0;
  return fact[n] * inv[r] % MOD * inv[n - r] % MOD;
}
void Solve(long long int n, long long int k) {
  g = 0;
  for (long long int i = 0, pos = 1; i <= k; i++, pos = MOD - pos) {
    g += pos * C(k, i) % MOD * pw(k - i, n, MOD) % MOD;
    g %= MOD;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    sum = (sum + A[i]) % MOD;
  }
  fact[0] = inv[0] = 1;
  for (long long int i = 1; i < N; i++) {
    fact[i] = i * fact[i - 1] % MOD;
    inv[i] = pw(fact[i], MOD - 2, MOD);
  }
  Solve(n - 1, k);
  ans = g * sum % MOD * (n - 1) % MOD;
  Solve(n, k);
  ans = (ans + g * sum % MOD) % MOD;
  ans = ans * inv[k] % MOD;
  cout << ans << '\n';
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 int pw(long long int a, long long int b, long long int mod) {
  if (!b) return 1;
  if (b & 1) return a * pw(a * a % mod, b / 2, mod) % mod;
  return pw(a * a % mod, b / 2, mod) % mod;
}
const long long int N = 2e5 + 10;
const long long int MOD = 1e9 + 7;
const long long int INF = 1e18;
long long int n, k, A[N], sum, g, fact[N], inv[N], ans;
inline long long int C(long long int n, long long int r) {
  if (r < 0 || n < r) return 0;
  return fact[n] * inv[r] % MOD * inv[n - r] % MOD;
}
void Solve(long long int n, long long int k) {
  g = 0;
  for (long long int i = 0, pos = 1; i <= k; i++, pos = MOD - pos) {
    g += pos * C(k, i) % MOD * pw(k - i, n, MOD) % MOD;
    g %= MOD;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    sum = (sum + A[i]) % MOD;
  }
  fact[0] = inv[0] = 1;
  for (long long int i = 1; i < N; i++) {
    fact[i] = i * fact[i - 1] % MOD;
    inv[i] = pw(fact[i], MOD - 2, MOD);
  }
  Solve(n - 1, k);
  ans = g * sum % MOD * (n - 1) % MOD;
  Solve(n, k);
  ans = (ans + g * sum % MOD) % MOD;
  ans = ans * inv[k] % MOD;
  cout << ans << '\n';
  return 0;
}

```