#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const int maxn = 200010;
long long sum, n, k, w;
long long f[maxn], inf[maxn];
long long p(long long a, long long n) {
  if (!~a) return n & 1 ? -1 : 1;
  long long b = 1;
  while (n) {
    if (n & 1) (b *= a) %= mod;
    (a *= a) %= mod;
    n >>= 1;
  }
  return b;
}
long long c(long long n, long long k) {
  if (k > n || n < 0) return 0;
  if (k == 0 || n == k) return 1;
  return f[n] * inf[k] % mod * inf[n - k] % mod;
}
long long stirling(long long n, long long k) {
  long long ret = 0;
  for (long long j = 0; j <= k; j++)
    (ret += p(-1, k + j) * c(k, j) % mod * p(j, n) % mod) %= mod;
  return ret * inf[k] % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> w;
    (sum += w) %= mod;
  }
  f[0] = inf[0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i] = f[i - 1] * i % mod;
    inf[i] = inf[i - 1] * p(i, mod - 2) % mod;
  }
  long long coe1 = stirling(n, k) % mod;
  long long coe2 = stirling(n - 1, k) * (n - 1) % mod;
  cout << ((coe1 + coe2 + mod) % mod * sum % mod + mod) % mod << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 mod = 1000000007;
const int maxn = 200010;
long long sum, n, k, w;
long long f[maxn], inf[maxn];
long long p(long long a, long long n) {
  if (!~a) return n & 1 ? -1 : 1;
  long long b = 1;
  while (n) {
    if (n & 1) (b *= a) %= mod;
    (a *= a) %= mod;
    n >>= 1;
  }
  return b;
}
long long c(long long n, long long k) {
  if (k > n || n < 0) return 0;
  if (k == 0 || n == k) return 1;
  return f[n] * inf[k] % mod * inf[n - k] % mod;
}
long long stirling(long long n, long long k) {
  long long ret = 0;
  for (long long j = 0; j <= k; j++)
    (ret += p(-1, k + j) * c(k, j) % mod * p(j, n) % mod) %= mod;
  return ret * inf[k] % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> w;
    (sum += w) %= mod;
  }
  f[0] = inf[0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i] = f[i - 1] * i % mod;
    inf[i] = inf[i - 1] * p(i, mod - 2) % mod;
  }
  long long coe1 = stirling(n, k) % mod;
  long long coe2 = stirling(n - 1, k) * (n - 1) % mod;
  cout << ((coe1 + coe2 + mod) % mod * sum % mod + mod) % mod << endl;
  return 0;
}

```