#include <bits/stdc++.h>
using namespace std;
const long long N = 2 * 1000 * 100 + 10, mod = 1LL * 1000 * 1000 * 1000 + 7;
long long fact[N], inv_fact[N], n, a[N], K, s[N], all[N];
long long Pow(long long a, long long b) {
  return b ? 1LL * (b & 1 ? a : 1) * Pow(1LL * a * a % mod, b / 2) % mod : 1;
}
void pre() {
  fact[0] = 1;
  for (long long i = 1; i < N; i++) fact[i] = 1LL * fact[i - 1] * i % mod;
  inv_fact[N - 1] = Pow(fact[N - 1], mod - 2);
  for (long long i = N - 2; i >= 0; i--)
    inv_fact[i] = 1LL * inv_fact[i + 1] * (i + 1) % mod;
}
long long C(long long r, long long n) {
  return 1LL * fact[n] * inv_fact[r] % mod * inv_fact[n - r] % mod;
}
long long F(long long n, long long k) {
  if (n == k) return 1;
  if (k > n) return 0;
  if (k == 1) return 1;
  long long res = 0;
  for (long long j = 0; j <= k; j++) {
    (res += 1LL * ((j + k) & 1 ? -1 : 1) * C(j, k) % mod * Pow(j, n) % mod) %=
        mod;
    (res %= mod);
    (res = res + mod) % mod;
  }
  return 1LL * res * inv_fact[k] % mod;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> K;
  pre();
  for (long long i = 1; i <= n; i++)
    cin >> a[i], s[i] = (s[i - 1] + a[i]) % mod;
  long long C = (1LL * F(n - 1, K) * (n - 1) % mod + F(n, K)) % mod;
  cout << 1LL * s[n] * C % mod << '\n';
  return 0;
}

### Prompt

Create a solution in Cpp for 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 N = 2 * 1000 * 100 + 10, mod = 1LL * 1000 * 1000 * 1000 + 7;
long long fact[N], inv_fact[N], n, a[N], K, s[N], all[N];
long long Pow(long long a, long long b) {
  return b ? 1LL * (b & 1 ? a : 1) * Pow(1LL * a * a % mod, b / 2) % mod : 1;
}
void pre() {
  fact[0] = 1;
  for (long long i = 1; i < N; i++) fact[i] = 1LL * fact[i - 1] * i % mod;
  inv_fact[N - 1] = Pow(fact[N - 1], mod - 2);
  for (long long i = N - 2; i >= 0; i--)
    inv_fact[i] = 1LL * inv_fact[i + 1] * (i + 1) % mod;
}
long long C(long long r, long long n) {
  return 1LL * fact[n] * inv_fact[r] % mod * inv_fact[n - r] % mod;
}
long long F(long long n, long long k) {
  if (n == k) return 1;
  if (k > n) return 0;
  if (k == 1) return 1;
  long long res = 0;
  for (long long j = 0; j <= k; j++) {
    (res += 1LL * ((j + k) & 1 ? -1 : 1) * C(j, k) % mod * Pow(j, n) % mod) %=
        mod;
    (res %= mod);
    (res = res + mod) % mod;
  }
  return 1LL * res * inv_fact[k] % mod;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> K;
  pre();
  for (long long i = 1; i <= n; i++)
    cin >> a[i], s[i] = (s[i - 1] + a[i]) % mod;
  long long C = (1LL * F(n - 1, K) * (n - 1) % mod + F(n, K)) % mod;
  cout << 1LL * s[n] * C % mod << '\n';
  return 0;
}

```