#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
long long Mul(long long a, long long b) { return a * b % MOD; }
void Add(long long &a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
long long gt[1001000], rev_gt[1001000];
long long POW(long long a, long long k) {
  if (k == 0) return 1;
  long long temp = POW(a, k >> 1);
  temp = Mul(temp, temp);
  if (k % 2) temp = Mul(temp, a);
  return temp;
}
long long C(int k, int n) { return Mul(gt[n], Mul(rev_gt[k], rev_gt[n - k])); }
long long f(int n, int k) {
  if (n < k) return 0;
  int hs = 1;
  if (k % 2) hs = -1;
  long long ans = 0;
  for (int i = (1); i <= (k); ++i) {
    hs = -1 * hs;
    long long now = Mul(C(i, k), POW(i, n));
    if (hs == -1) now = MOD - now;
    Add(ans, now);
  }
  return Mul(ans, rev_gt[k]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  gt[0] = 1;
  for (int i = (1); i <= (n); ++i) gt[i] = Mul(gt[i - 1], i);
  rev_gt[n] = POW(gt[n], MOD - 2);
  for (int i = (n - 1); i >= (0); --i) rev_gt[i] = Mul(rev_gt[i + 1], i + 1);
  long long sum = 0;
  for (int i = (1); i <= (n); ++i) {
    int x;
    cin >> x;
    Add(sum, x);
  }
  long long ans = f(n, k);
  Add(ans, Mul(n - 1, f(n - 1, k)));
  cout << Mul(ans, sum);
}

### Prompt

In cpp, your task is to solve 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 long long MOD = 1e9 + 7;
long long Mul(long long a, long long b) { return a * b % MOD; }
void Add(long long &a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
long long gt[1001000], rev_gt[1001000];
long long POW(long long a, long long k) {
  if (k == 0) return 1;
  long long temp = POW(a, k >> 1);
  temp = Mul(temp, temp);
  if (k % 2) temp = Mul(temp, a);
  return temp;
}
long long C(int k, int n) { return Mul(gt[n], Mul(rev_gt[k], rev_gt[n - k])); }
long long f(int n, int k) {
  if (n < k) return 0;
  int hs = 1;
  if (k % 2) hs = -1;
  long long ans = 0;
  for (int i = (1); i <= (k); ++i) {
    hs = -1 * hs;
    long long now = Mul(C(i, k), POW(i, n));
    if (hs == -1) now = MOD - now;
    Add(ans, now);
  }
  return Mul(ans, rev_gt[k]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  gt[0] = 1;
  for (int i = (1); i <= (n); ++i) gt[i] = Mul(gt[i - 1], i);
  rev_gt[n] = POW(gt[n], MOD - 2);
  for (int i = (n - 1); i >= (0); --i) rev_gt[i] = Mul(rev_gt[i + 1], i + 1);
  long long sum = 0;
  for (int i = (1); i <= (n); ++i) {
    int x;
    cin >> x;
    Add(sum, x);
  }
  long long ans = f(n, k);
  Add(ans, Mul(n - 1, f(n - 1, k)));
  cout << Mul(ans, sum);
}

```