#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 123;
const long long mod = 1e9 + 7;
int n, k;
long long h[N], res = 0, sum = 0;
long long hpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long a = hpow(x, y / 2);
  a = (a * a) % mod;
  return y & 1 ? (a * x) % mod : a;
}
long long rev(long long x) { return hpow(x, mod - 2); }
long long combination(long long k, long long n) {
  if (k < 0 || k > n) return 0;
  return ((h[n] * rev(h[k])) % mod * rev(h[n - k])) % mod;
}
long long S(int n, int m) {
  long long cnt = 0, z = -1;
  for (int i = 0; i <= m; ++i) {
    z = -z;
    long long x = (z * combination(i, m) % mod) * hpow(m - i, n) % mod;
    cnt = (cnt + x) % mod;
  }
  return rev(h[k]) * cnt % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    long long x;
    cin >> x;
    sum += x;
  }
  sum %= mod;
  if (k == 1) {
    cout << sum * n % mod;
    return 0;
  }
  h[0] = 1;
  for (int i = 1; i <= n; ++i) h[i] = (h[i - 1] * i) % mod;
  long long res = (S(n, k) + S(n - 1, k) * (n - 1) % mod) % mod;
  res = res * sum % mod;
  if (res < 0) res += mod;
  cout << res;
  return 0;
}

### Prompt

Generate 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 N = 2e5 + 123;
const long long mod = 1e9 + 7;
int n, k;
long long h[N], res = 0, sum = 0;
long long hpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long a = hpow(x, y / 2);
  a = (a * a) % mod;
  return y & 1 ? (a * x) % mod : a;
}
long long rev(long long x) { return hpow(x, mod - 2); }
long long combination(long long k, long long n) {
  if (k < 0 || k > n) return 0;
  return ((h[n] * rev(h[k])) % mod * rev(h[n - k])) % mod;
}
long long S(int n, int m) {
  long long cnt = 0, z = -1;
  for (int i = 0; i <= m; ++i) {
    z = -z;
    long long x = (z * combination(i, m) % mod) * hpow(m - i, n) % mod;
    cnt = (cnt + x) % mod;
  }
  return rev(h[k]) * cnt % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    long long x;
    cin >> x;
    sum += x;
  }
  sum %= mod;
  if (k == 1) {
    cout << sum * n % mod;
    return 0;
  }
  h[0] = 1;
  for (int i = 1; i <= n; ++i) h[i] = (h[i - 1] * i) % mod;
  long long res = (S(n, k) + S(n - 1, k) * (n - 1) % mod) % mod;
  res = res * sum % mod;
  if (res < 0) res += mod;
  cout << res;
  return 0;
}

```