#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
int fac[N], inv[N], ifac[N], a[N];
int add(int a, int b) {
  if (a + b >= mod) return a + b - mod;
  return a + b;
}
int mul(int a, int b) { return (long long)a * b % mod; }
long long qpow(long long a, long long b) {
  if (a < 0) return 0;
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a % mod;
    }
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
void init(int n) {
  fac[0] = inv[0] = ifac[0] = 1;
  for (int i = (1); i < (n + 1); i++) fac[i] = mul(fac[i - 1], i);
  ifac[n] = qpow(fac[n], mod - 2);
  for (int i = (n)-1; i >= (1); i--) ifac[i] = mul(ifac[i + 1], i + 1);
  for (int i = (1); i < (n + 1); i++) inv[i] = mul(ifac[i], fac[i - 1]);
}
int get_C(int a, int b) { return mul(mul(fac[a], ifac[a - b]), ifac[b]); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init(1e6);
  int n, k;
  cin >> n >> k;
  long long sum = 0;
  for (int i = (1); i < (n + 1); i++) {
    cin >> a[i];
    sum += a[i];
  }
  sum %= mod;
  if (k == 1) {
    cout << (long long)n * sum % mod << "\n";
    return 0;
  }
  int ans = 0;
  for (int i = (0); i < (k); i++) {
    int tmp = ((long long)ifac[k - 1] * get_C(k - 1, i) % mod *
                   qpow(k - i, n - 2) % mod * (n - 1) % mod +
               (long long)ifac[k - 1] * get_C(k - 1, i) % mod *
                   qpow(k - i, n - 1) % mod) %
              mod;
    if (i % 2 == 1)
      ans = (ans - tmp + mod) % mod;
    else
      ans = (ans + tmp) % mod;
  }
  cout << (long long)ans * sum % mod;
  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;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
int fac[N], inv[N], ifac[N], a[N];
int add(int a, int b) {
  if (a + b >= mod) return a + b - mod;
  return a + b;
}
int mul(int a, int b) { return (long long)a * b % mod; }
long long qpow(long long a, long long b) {
  if (a < 0) return 0;
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a % mod;
    }
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
void init(int n) {
  fac[0] = inv[0] = ifac[0] = 1;
  for (int i = (1); i < (n + 1); i++) fac[i] = mul(fac[i - 1], i);
  ifac[n] = qpow(fac[n], mod - 2);
  for (int i = (n)-1; i >= (1); i--) ifac[i] = mul(ifac[i + 1], i + 1);
  for (int i = (1); i < (n + 1); i++) inv[i] = mul(ifac[i], fac[i - 1]);
}
int get_C(int a, int b) { return mul(mul(fac[a], ifac[a - b]), ifac[b]); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init(1e6);
  int n, k;
  cin >> n >> k;
  long long sum = 0;
  for (int i = (1); i < (n + 1); i++) {
    cin >> a[i];
    sum += a[i];
  }
  sum %= mod;
  if (k == 1) {
    cout << (long long)n * sum % mod << "\n";
    return 0;
  }
  int ans = 0;
  for (int i = (0); i < (k); i++) {
    int tmp = ((long long)ifac[k - 1] * get_C(k - 1, i) % mod *
                   qpow(k - i, n - 2) % mod * (n - 1) % mod +
               (long long)ifac[k - 1] * get_C(k - 1, i) % mod *
                   qpow(k - i, n - 1) % mod) %
              mod;
    if (i % 2 == 1)
      ans = (ans - tmp + mod) % mod;
    else
      ans = (ans + tmp) % mod;
  }
  cout << (long long)ans * sum % mod;
  return 0;
}

```