#include <bits/stdc++.h>
using namespace std;
const long long N = 200005;
const long long MOD = 1000000007;
long long w[N], pre[N], inv[N], sum = 0, ans = 0;
long long qpow(long long a, long long n) {
  long long ans = 1, tmp = a;
  while (n) {
    if (n & 1) ans = ans * tmp % MOD;
    tmp = tmp * tmp % MOD;
    n >>= 1;
  }
  return ans;
}
long long C(long long n, long long m) {
  return pre[n] * inv[m] % MOD * inv[n - m] % MOD;
}
void init() {
  pre[0] = 1;
  for (long long i = 1; i < N; i++) pre[i] = pre[i - 1] * i % MOD;
  inv[N - 1] = qpow(pre[N - 1], MOD - 2);
  for (long long i = N - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % MOD;
}
signed main() {
  init();
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> w[i];
    sum = (sum + w[i]) % MOD;
  }
  if (n == 1) {
    cout << sum << endl;
    return 0;
  }
  for (long long i = 0, d = 1; i < k; i++, d = MOD - d) {
    long long ret = d * inv[i] % MOD * inv[k - 1 - i] % MOD *
                    qpow(k - i, n - 2) % MOD * (k - i + n - 1) % MOD;
    ans = (ans + ret) % MOD;
  }
  cout << ans * sum % MOD << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 long long N = 200005;
const long long MOD = 1000000007;
long long w[N], pre[N], inv[N], sum = 0, ans = 0;
long long qpow(long long a, long long n) {
  long long ans = 1, tmp = a;
  while (n) {
    if (n & 1) ans = ans * tmp % MOD;
    tmp = tmp * tmp % MOD;
    n >>= 1;
  }
  return ans;
}
long long C(long long n, long long m) {
  return pre[n] * inv[m] % MOD * inv[n - m] % MOD;
}
void init() {
  pre[0] = 1;
  for (long long i = 1; i < N; i++) pre[i] = pre[i - 1] * i % MOD;
  inv[N - 1] = qpow(pre[N - 1], MOD - 2);
  for (long long i = N - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % MOD;
}
signed main() {
  init();
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> w[i];
    sum = (sum + w[i]) % MOD;
  }
  if (n == 1) {
    cout << sum << endl;
    return 0;
  }
  for (long long i = 0, d = 1; i < k; i++, d = MOD - d) {
    long long ret = d * inv[i] % MOD * inv[k - 1 - i] % MOD *
                    qpow(k - i, n - 2) % MOD * (k - i + n - 1) % MOD;
    ans = (ans + ret) % MOD;
  }
  cout << ans * sum % MOD << endl;
  return 0;
}

```