#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
using namespace std;
const long long N = 3e5 + 100, OO = 1e9 + 7, T = 10, M = 1e9 + 7, P = 6151,
                SQ = 300, lg = 30;
long long fac[N], inv[N];
long long pw(long long x, long long y) {
  if (y == 0) return 1;
  long long cnt = pw(x, y / 2);
  cnt = (cnt * cnt) % M;
  cnt = (cnt * (y % 2 == 1 ? x : 1)) % M;
  return cnt;
}
void prepro() {
  fac[0] = 1;
  for (long long i = 1; i <= N - lg; i++) fac[i] = (fac[i - 1] * i) % M;
  inv[N - lg] = pw(fac[N - lg], M - 2);
  for (long long i = N - lg - 1; i > -1; i--)
    inv[i] = (inv[i + 1] * (i + 1)) % M;
}
long long c(long long x, long long y) {
  return ((fac[x] * inv[y] % M) * inv[x - y]) % M;
}
long long get(long long n, long long k) {
  if (n < k) return 0;
  long long cnt = inv[k], sum = 0;
  for (long long i = 0; i <= k; i++) {
    if (i % 2 == 0)
      sum = (sum + c(k, i) * pw(k - i, n)) % M;
    else
      sum = (sum - c(k, i) * pw(k - i, n)) % M;
  }
  sum = (sum + M) % M;
  cnt = (cnt * sum) % M;
  return cnt;
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  prepro();
  long long n, k;
  cin >> n >> k;
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    sum = (sum + x) % M;
  }
  long long ans = (get(n, k) + ((n - 1) * get(n - 1, k)) % M) % M;
  ans = (ans * sum) % M;
  cout << ans << endl;
  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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
using namespace std;
const long long N = 3e5 + 100, OO = 1e9 + 7, T = 10, M = 1e9 + 7, P = 6151,
                SQ = 300, lg = 30;
long long fac[N], inv[N];
long long pw(long long x, long long y) {
  if (y == 0) return 1;
  long long cnt = pw(x, y / 2);
  cnt = (cnt * cnt) % M;
  cnt = (cnt * (y % 2 == 1 ? x : 1)) % M;
  return cnt;
}
void prepro() {
  fac[0] = 1;
  for (long long i = 1; i <= N - lg; i++) fac[i] = (fac[i - 1] * i) % M;
  inv[N - lg] = pw(fac[N - lg], M - 2);
  for (long long i = N - lg - 1; i > -1; i--)
    inv[i] = (inv[i + 1] * (i + 1)) % M;
}
long long c(long long x, long long y) {
  return ((fac[x] * inv[y] % M) * inv[x - y]) % M;
}
long long get(long long n, long long k) {
  if (n < k) return 0;
  long long cnt = inv[k], sum = 0;
  for (long long i = 0; i <= k; i++) {
    if (i % 2 == 0)
      sum = (sum + c(k, i) * pw(k - i, n)) % M;
    else
      sum = (sum - c(k, i) * pw(k - i, n)) % M;
  }
  sum = (sum + M) % M;
  cnt = (cnt * sum) % M;
  return cnt;
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  prepro();
  long long n, k;
  cin >> n >> k;
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    sum = (sum + x) % M;
  }
  long long ans = (get(n, k) + ((n - 1) * get(n - 1, k)) % M) % M;
  ans = (ans * sum) % M;
  cout << ans << endl;
  return 0;
}

```