#include <bits/stdc++.h>
using namespace std;
clock_t start;
void Time(bool timeIt) {
  if (!timeIt) return;
  clock_t end = clock();
  double elapsed_time = ((double)end - (double)start) / (double)CLOCKS_PER_SEC;
  fprintf(stderr, "Time elapsed = %0.4lf\n", elapsed_time);
}
const long long mod = 1e9 + 7;
long long f[200010], inv[200010];
long long exp(long long b, long long e) {
  long long ret = 1;
  while (e > 0) {
    if (e & 1) ret = (ret * b) % mod;
    e >>= 1;
    b = (b * b) % mod;
  }
  return ret;
}
long long stirling(int n, int k) {
  if (n < k) return 0;
  long long ret = 0;
  long long m = (k & 1) ? -1 : 1;
  for (int i = 0; i <= k; ++i) {
    long long c = (f[k] * ((inv[k - i] * inv[i]) % mod)) % mod;
    ret += m * c * exp(i, n) % mod;
    if (ret >= mod) ret -= mod;
    if (ret < 0) ret += mod;
    m *= -1;
  }
  ret = (ret * inv[k]) % mod;
  return ret;
}
int main() {
  start = clock();
  f[0] = inv[0] = 1;
  for (int i = 1; i < 200010; ++i) {
    f[i] = (1ll * f[i - 1] * i) % mod;
  }
  inv[200010 - 1] = exp(f[200010 - 1], mod - 2);
  for (int i = 200010 - 2; i > 0; --i) {
    inv[i] = (1ll * inv[i + 1] * (i + 1)) % mod;
  }
  int n, k;
  scanf("%d %d", &n, &k);
  long long sum = 0;
  for (int i = 0; i < n; ++i) {
    long long w;
    scanf("%lld", &w);
    sum += w;
    if (sum >= mod) sum -= mod;
  }
  long long mult =
      (stirling(n, k) + (1ll * (n - 1) * stirling(n - 1, k)) % mod) % mod;
  printf("%lld\n", (sum * mult) % mod);
  Time(true);
  return 0;
}

### Prompt

Please formulate 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;
clock_t start;
void Time(bool timeIt) {
  if (!timeIt) return;
  clock_t end = clock();
  double elapsed_time = ((double)end - (double)start) / (double)CLOCKS_PER_SEC;
  fprintf(stderr, "Time elapsed = %0.4lf\n", elapsed_time);
}
const long long mod = 1e9 + 7;
long long f[200010], inv[200010];
long long exp(long long b, long long e) {
  long long ret = 1;
  while (e > 0) {
    if (e & 1) ret = (ret * b) % mod;
    e >>= 1;
    b = (b * b) % mod;
  }
  return ret;
}
long long stirling(int n, int k) {
  if (n < k) return 0;
  long long ret = 0;
  long long m = (k & 1) ? -1 : 1;
  for (int i = 0; i <= k; ++i) {
    long long c = (f[k] * ((inv[k - i] * inv[i]) % mod)) % mod;
    ret += m * c * exp(i, n) % mod;
    if (ret >= mod) ret -= mod;
    if (ret < 0) ret += mod;
    m *= -1;
  }
  ret = (ret * inv[k]) % mod;
  return ret;
}
int main() {
  start = clock();
  f[0] = inv[0] = 1;
  for (int i = 1; i < 200010; ++i) {
    f[i] = (1ll * f[i - 1] * i) % mod;
  }
  inv[200010 - 1] = exp(f[200010 - 1], mod - 2);
  for (int i = 200010 - 2; i > 0; --i) {
    inv[i] = (1ll * inv[i + 1] * (i + 1)) % mod;
  }
  int n, k;
  scanf("%d %d", &n, &k);
  long long sum = 0;
  for (int i = 0; i < n; ++i) {
    long long w;
    scanf("%lld", &w);
    sum += w;
    if (sum >= mod) sum -= mod;
  }
  long long mult =
      (stirling(n, k) + (1ll * (n - 1) * stirling(n - 1, k)) % mod) % mod;
  printf("%lld\n", (sum * mult) % mod);
  Time(true);
  return 0;
}

```