#include <bits/stdc++.h>
using namespace std;
const long long Mod = 1e9 + 7;
int w[200010];
long long fac[200010], inv[200010];
long long pw(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = res * x % Mod;
    }
    x = x * x % Mod;
    y >>= 1;
  }
  return res;
}
long long c(int n, int m) {
  if (m < 0 || m > n) {
    return 0;
  }
  return fac[n] * inv[n - m] % Mod * inv[m] % Mod;
}
long long cal(int n, int k) {
  long long res = 0;
  for (int i = 0; i <= k; i++) {
    long long v = c(k, i) * pw(k - i, n) % Mod;
    if (i & 1) {
      v = (Mod - v) % Mod;
    }
    res = (res + v) % Mod;
  }
  res = res * inv[k] % Mod;
  return res;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % Mod;
  }
  inv[n] = pw(fac[n], Mod - 2);
  for (int i = n - 1; i >= 0; i--) {
    inv[i] = inv[i + 1] * (i + 1) % Mod;
  }
  long long ans = 0, s = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    s = (s + w[i]) % Mod;
  }
  ans = (cal(n, k) + 1LL * (n - 1) * cal(n - 1, k) % Mod) % Mod;
  ans = ans * s % Mod;
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const long long Mod = 1e9 + 7;
int w[200010];
long long fac[200010], inv[200010];
long long pw(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = res * x % Mod;
    }
    x = x * x % Mod;
    y >>= 1;
  }
  return res;
}
long long c(int n, int m) {
  if (m < 0 || m > n) {
    return 0;
  }
  return fac[n] * inv[n - m] % Mod * inv[m] % Mod;
}
long long cal(int n, int k) {
  long long res = 0;
  for (int i = 0; i <= k; i++) {
    long long v = c(k, i) * pw(k - i, n) % Mod;
    if (i & 1) {
      v = (Mod - v) % Mod;
    }
    res = (res + v) % Mod;
  }
  res = res * inv[k] % Mod;
  return res;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % Mod;
  }
  inv[n] = pw(fac[n], Mod - 2);
  for (int i = n - 1; i >= 0; i--) {
    inv[i] = inv[i + 1] * (i + 1) % Mod;
  }
  long long ans = 0, s = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    s = (s + w[i]) % Mod;
  }
  ans = (cal(n, k) + 1LL * (n - 1) * cal(n - 1, k) % Mod) % Mod;
  ans = ans * s % Mod;
  cout << ans << endl;
  return 0;
}

```