#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
const int N = 2e5 + 10;
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
const long long inf = 8e18;
const int LOG = 22;
long long pw(long long a, long long b, long long M) {
  return (!b ? 1
             : (b & 1 ? (a * pw(a * a % M, b / 2, M)) % M
                      : pw(a * a % M, b / 2, M)));
}
long long sum, first[N], I[N];
long long E(long long r, long long n) {
  if (r > n || r < 0) return 0ll;
  return first[n] * I[r] % mod * I[n - r] % mod;
}
long long sterling(int n, int k) {
  long long ans = 0;
  for (int j = 0; j <= k; j++) {
    long long z = ((k - j) & 1 ? -1 : 1);
    ans = (ans + z * (E(j, k) * pw(j, n, mod) % mod) + 2 * mod) % mod;
  }
  return ans * I[k] % mod;
}
int main() {
  first[0] = 1;
  for (int i = 1; i < N; i++) {
    first[i] = first[i - 1] * i % mod;
  }
  I[N - 1] = pw(first[N - 1], mod - 2, mod);
  for (int i = N - 2; ~i; i--) {
    I[i] = I[i + 1] * (i + 1) % mod;
  }
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int cu;
    scanf("%d", &cu);
    sum += cu;
  }
  sum %= mod;
  long long tot =
      sum * (sterling(n, k) + (sterling(n - 1, k) * (n - 1) % mod)) % mod;
  printf("%lld\n", tot);
  return 0;
}

### 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>
#pragma GCC optimize("O2")
using namespace std;
const int N = 2e5 + 10;
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
const long long inf = 8e18;
const int LOG = 22;
long long pw(long long a, long long b, long long M) {
  return (!b ? 1
             : (b & 1 ? (a * pw(a * a % M, b / 2, M)) % M
                      : pw(a * a % M, b / 2, M)));
}
long long sum, first[N], I[N];
long long E(long long r, long long n) {
  if (r > n || r < 0) return 0ll;
  return first[n] * I[r] % mod * I[n - r] % mod;
}
long long sterling(int n, int k) {
  long long ans = 0;
  for (int j = 0; j <= k; j++) {
    long long z = ((k - j) & 1 ? -1 : 1);
    ans = (ans + z * (E(j, k) * pw(j, n, mod) % mod) + 2 * mod) % mod;
  }
  return ans * I[k] % mod;
}
int main() {
  first[0] = 1;
  for (int i = 1; i < N; i++) {
    first[i] = first[i - 1] * i % mod;
  }
  I[N - 1] = pw(first[N - 1], mod - 2, mod);
  for (int i = N - 2; ~i; i--) {
    I[i] = I[i + 1] * (i + 1) % mod;
  }
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int cu;
    scanf("%d", &cu);
    sum += cu;
  }
  sum %= mod;
  long long tot =
      sum * (sterling(n, k) + (sterling(n - 1, k) * (n - 1) % mod)) % mod;
  printf("%lld\n", tot);
  return 0;
}

```