#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
const int maxn = 3e5;
long long int fac[maxn] = {}, inv[maxn] = {};
long long int power(long long int x, long long int y) {
  long long int res = 1, tem = x;
  for (; y; y >>= 1) {
    if (y & 1) res = (res % mod) * (tem % mod) % mod;
    tem = (tem % mod) * (tem % mod) % mod;
  }
  return res;
}
long long int C(int n, int m) {
  if (m == 0 || n == m) return 1;
  long long res = 1;
  res = res * fac[n] % mod * inv[m] % mod * inv[n - m] % mod;
  return res;
}
long long int S(long long int n, long long int m) {
  long long int ans = 0;
  for (int i = 0; i <= m; i++) {
    long long int t = 1ll * inv[i] * inv[m - i] % mod * power(m - i, n) % mod;
    if (i & 1)
      (ans -= t) %= mod;
    else
      (ans += t) %= mod;
  }
  return ans;
}
int main() {
  long long int ans1 = 0, n, k;
  scanf("%lld%lld", &n, &k);
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= k; i++) inv[i] = -1ll * mod / i * inv[mod % i] % mod;
  for (int i = 1; i <= k; i++) inv[i] = 1ll * inv[i - 1] * inv[i] % mod;
  for (int i = 1; i <= n; i++) {
    long long int w;
    scanf("%lld", &w);
    ans1 = (ans1 + w) % mod;
  }
  long long int tem1 = 1, tem2, ans = 0;
  for (long long int i = 2; i <= k; i++) {
    tem1 = (tem1 * i) % mod;
    if (i == k - 1) tem2 = tem1;
  }
  ans = (S(n - 1, k - 1) % mod + 1ll * (n + k - 1) * S(n - 1, k) % mod) % mod;
  printf("%lld\n", (ans * ans1 % mod + mod) % mod);
  return 0;
}

### Prompt

Generate a CPP solution to 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 int mod = 1e9 + 7;
const int maxn = 3e5;
long long int fac[maxn] = {}, inv[maxn] = {};
long long int power(long long int x, long long int y) {
  long long int res = 1, tem = x;
  for (; y; y >>= 1) {
    if (y & 1) res = (res % mod) * (tem % mod) % mod;
    tem = (tem % mod) * (tem % mod) % mod;
  }
  return res;
}
long long int C(int n, int m) {
  if (m == 0 || n == m) return 1;
  long long res = 1;
  res = res * fac[n] % mod * inv[m] % mod * inv[n - m] % mod;
  return res;
}
long long int S(long long int n, long long int m) {
  long long int ans = 0;
  for (int i = 0; i <= m; i++) {
    long long int t = 1ll * inv[i] * inv[m - i] % mod * power(m - i, n) % mod;
    if (i & 1)
      (ans -= t) %= mod;
    else
      (ans += t) %= mod;
  }
  return ans;
}
int main() {
  long long int ans1 = 0, n, k;
  scanf("%lld%lld", &n, &k);
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= k; i++) inv[i] = -1ll * mod / i * inv[mod % i] % mod;
  for (int i = 1; i <= k; i++) inv[i] = 1ll * inv[i - 1] * inv[i] % mod;
  for (int i = 1; i <= n; i++) {
    long long int w;
    scanf("%lld", &w);
    ans1 = (ans1 + w) % mod;
  }
  long long int tem1 = 1, tem2, ans = 0;
  for (long long int i = 2; i <= k; i++) {
    tem1 = (tem1 * i) % mod;
    if (i == k - 1) tem2 = tem1;
  }
  ans = (S(n - 1, k - 1) % mod + 1ll * (n + k - 1) * S(n - 1, k) % mod) % mod;
  printf("%lld\n", (ans * ans1 % mod + mod) % mod);
  return 0;
}

```