#include <bits/stdc++.h>
using namespace std;
const int p = 1e9 + 7;
const int maxn = 2e5 + 5;
int fac[maxn], inv[maxn];
int ksm(int a, int b, int p) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % p;
    b >>= 1, a = 1ll * a * a % p;
  }
  return ans;
}
int s(int n, int k) {
  int ans = 0;
  for (int i = 0; i <= k; ++i)
    ans = 1ll *
          (ans + 1ll * ((i & 1) ? (p - 1) : 1) * inv[i] % p * ksm(k - i, n, p) %
                     p * inv[k - i] % p) %
          p;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n, k, x;
  cin >> n >> k;
  int sum = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> x;
    sum += x;
    sum %= p;
  }
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % p;
  inv[n] = ksm(fac[n], p - 2, p);
  for (int i = n - 1; i >= 1; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % p;
  cout << 1ll * sum * (s(n, k) + 1ll * (n - 1) * s(n - 1, k) % p) % p << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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 int p = 1e9 + 7;
const int maxn = 2e5 + 5;
int fac[maxn], inv[maxn];
int ksm(int a, int b, int p) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % p;
    b >>= 1, a = 1ll * a * a % p;
  }
  return ans;
}
int s(int n, int k) {
  int ans = 0;
  for (int i = 0; i <= k; ++i)
    ans = 1ll *
          (ans + 1ll * ((i & 1) ? (p - 1) : 1) * inv[i] % p * ksm(k - i, n, p) %
                     p * inv[k - i] % p) %
          p;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n, k, x;
  cin >> n >> k;
  int sum = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> x;
    sum += x;
    sum %= p;
  }
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % p;
  inv[n] = ksm(fac[n], p - 2, p);
  for (int i = n - 1; i >= 1; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % p;
  cout << 1ll * sum * (s(n, k) + 1ll * (n - 1) * s(n - 1, k) % p) % p << '\n';
  return 0;
}

```