#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, mod = 1e9 + 7;
int n, k, sum, fc[N], ifc[N];
int ch(int x, int y) {
  if (y == 0 || x > y) return 0;
  return 1ll * fc[y] * ifc[x] % mod * ifc[y - x] % mod;
}
int power(int a, int b) {
  int rtr = 1;
  for (; b > 0; b /= 2, a = 1ll * a * a % mod)
    if (b & 1) rtr = 1ll * rtr * a % mod;
  return rtr;
}
int getAns(int num, int prt) {
  int ans = 0;
  for (int i = 1; i <= prt; i++) {
    int rtr = 1ll * ch(i, prt) * power(i, num) % mod;
    if (k % 2 == i % 2)
      ans = (ans + rtr) % mod;
    else
      ans = (ans - rtr + mod) % mod;
  }
  return 1ll * ans * ifc[prt] % mod;
}
void Pre() {
  fc[0] = ifc[0] = 1;
  for (int i = 1; i < N - 5; i++)
    fc[i] = 1ll * fc[i - 1] * i % mod, ifc[i] = power(fc[i], mod - 2);
  return;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  Pre();
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    sum = (sum + x) % mod;
  }
  if (k == 1) {
    cout << 1ll * n * sum % mod;
    return 0;
  }
  if (k == n) {
    cout << sum;
    return 0;
  }
  cout << (1ll * (getAns(n, k) + 1ll * (n - 1) * getAns(n - 1, k) % mod) % mod *
           sum % mod) %
              mod;
  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 int N = 2e5 + 10, mod = 1e9 + 7;
int n, k, sum, fc[N], ifc[N];
int ch(int x, int y) {
  if (y == 0 || x > y) return 0;
  return 1ll * fc[y] * ifc[x] % mod * ifc[y - x] % mod;
}
int power(int a, int b) {
  int rtr = 1;
  for (; b > 0; b /= 2, a = 1ll * a * a % mod)
    if (b & 1) rtr = 1ll * rtr * a % mod;
  return rtr;
}
int getAns(int num, int prt) {
  int ans = 0;
  for (int i = 1; i <= prt; i++) {
    int rtr = 1ll * ch(i, prt) * power(i, num) % mod;
    if (k % 2 == i % 2)
      ans = (ans + rtr) % mod;
    else
      ans = (ans - rtr + mod) % mod;
  }
  return 1ll * ans * ifc[prt] % mod;
}
void Pre() {
  fc[0] = ifc[0] = 1;
  for (int i = 1; i < N - 5; i++)
    fc[i] = 1ll * fc[i - 1] * i % mod, ifc[i] = power(fc[i], mod - 2);
  return;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  Pre();
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    sum = (sum + x) % mod;
  }
  if (k == 1) {
    cout << 1ll * n * sum % mod;
    return 0;
  }
  if (k == n) {
    cout << sum;
    return 0;
  }
  cout << (1ll * (getAns(n, k) + 1ll * (n - 1) * getAns(n - 1, k) % mod) % mod *
           sum % mod) %
              mod;
  return 0;
}

```