#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int MN = 200010;
int exp(int x, int n) {
  int ret = 1;
  while (n) {
    if (n & 1) ret = 1LL * ret * x % mod;
    x = 1LL * x * x % mod;
    n >>= 1;
  }
  return ret;
}
int inv(int x) { return exp(x, mod - 2); }
int fact[MN], invf[MN];
int comb(int n, int k) {
  if (n < k) return 0;
  return 1LL * fact[n] * invf[k] % mod * invf[n - k] % mod;
}
int S(int n, int k) {
  int ret = 0;
  for (int i = 0; i < k; i++) {
    ret += 1LL * comb(k, i) * (i % 2 ? mod - 1 : 1) % mod * exp(k - i, n) % mod;
    ret %= mod;
  }
  ret = 1LL * ret * invf[k] % mod;
  return ret;
}
int N, K;
int W[MN];
int main() {
  fact[0] = 1;
  for (int i = 1; i < MN; i++) {
    fact[i] = 1LL * fact[i - 1] * i % mod;
  }
  for (int i = 0; i < MN; i++) {
    invf[i] = inv(fact[i]);
  }
  scanf("%d %d", &N, &K);
  int sum = 0;
  for (int i = 0; i < N; i++) {
    scanf("%d", &W[i]);
    sum += W[i];
    sum %= mod;
  }
  int ans =
      1LL * ((S(N, K) + 1LL * (N - 1) * S(N - 1, K) % mod) % mod) * sum % mod;
  printf("%d", ans);
}

### 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 mod = 1e9 + 7;
const int MN = 200010;
int exp(int x, int n) {
  int ret = 1;
  while (n) {
    if (n & 1) ret = 1LL * ret * x % mod;
    x = 1LL * x * x % mod;
    n >>= 1;
  }
  return ret;
}
int inv(int x) { return exp(x, mod - 2); }
int fact[MN], invf[MN];
int comb(int n, int k) {
  if (n < k) return 0;
  return 1LL * fact[n] * invf[k] % mod * invf[n - k] % mod;
}
int S(int n, int k) {
  int ret = 0;
  for (int i = 0; i < k; i++) {
    ret += 1LL * comb(k, i) * (i % 2 ? mod - 1 : 1) % mod * exp(k - i, n) % mod;
    ret %= mod;
  }
  ret = 1LL * ret * invf[k] % mod;
  return ret;
}
int N, K;
int W[MN];
int main() {
  fact[0] = 1;
  for (int i = 1; i < MN; i++) {
    fact[i] = 1LL * fact[i - 1] * i % mod;
  }
  for (int i = 0; i < MN; i++) {
    invf[i] = inv(fact[i]);
  }
  scanf("%d %d", &N, &K);
  int sum = 0;
  for (int i = 0; i < N; i++) {
    scanf("%d", &W[i]);
    sum += W[i];
    sum %= mod;
  }
  int ans =
      1LL * ((S(N, K) + 1LL * (N - 1) * S(N - 1, K) % mod) % mod) * sum % mod;
  printf("%d", ans);
}

```