#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
long long n, k;
long long w[200010];
long long val1, val2;
long long fac[200010], ans, inv[200010];
inline long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    b >>= 1;
    a = a * a % 1000000007;
  }
  return ans;
}
inline long long C(long long n, long long m) {
  return fac[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
inline long long STL2(long long n, long long m) {
  long long sum = 0, c = 1;
  for (register long long i = 0; i <= m; i++) {
    sum = (sum + c * C(m, i) * ksm(m - i, n) % 1000000007) % 1000000007;
    c = -c;
  }
  return sum * inv[m] % 1000000007;
}
int main() {
  n = read(), k = read();
  for (register long long i = 1; i <= n; i++) w[i] = read();
  fac[0] = 1;
  inv[0] = inv[1] = 1;
  for (register long long i = 1; i <= n; i++)
    fac[i] = fac[i - 1] * i % 1000000007;
  for (register long long i = 2; i <= n; i++)
    inv[i] = inv[1000000007 % i] * (1000000007 - 1000000007 / i) % 1000000007;
  for (register long long i = 2; i <= n; i++)
    inv[i] = inv[i] * inv[i - 1] % 1000000007;
  val1 = STL2(n - 1, k), val2 = STL2(n, k);
  val1 = val1 * (n - 1) % 1000000007;
  for (register long long i = 1; i <= n; i++)
    ans = (ans + w[i] * (val1 + val2) % 1000000007) % 1000000007;
  printf("%lld\n", (ans + 1000000007) % 1000000007);
  return 0;
}

### Prompt

Your task is to create 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;
inline long long read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
long long n, k;
long long w[200010];
long long val1, val2;
long long fac[200010], ans, inv[200010];
inline long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    b >>= 1;
    a = a * a % 1000000007;
  }
  return ans;
}
inline long long C(long long n, long long m) {
  return fac[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
inline long long STL2(long long n, long long m) {
  long long sum = 0, c = 1;
  for (register long long i = 0; i <= m; i++) {
    sum = (sum + c * C(m, i) * ksm(m - i, n) % 1000000007) % 1000000007;
    c = -c;
  }
  return sum * inv[m] % 1000000007;
}
int main() {
  n = read(), k = read();
  for (register long long i = 1; i <= n; i++) w[i] = read();
  fac[0] = 1;
  inv[0] = inv[1] = 1;
  for (register long long i = 1; i <= n; i++)
    fac[i] = fac[i - 1] * i % 1000000007;
  for (register long long i = 2; i <= n; i++)
    inv[i] = inv[1000000007 % i] * (1000000007 - 1000000007 / i) % 1000000007;
  for (register long long i = 2; i <= n; i++)
    inv[i] = inv[i] * inv[i - 1] % 1000000007;
  val1 = STL2(n - 1, k), val2 = STL2(n, k);
  val1 = val1 * (n - 1) % 1000000007;
  for (register long long i = 1; i <= n; i++)
    ans = (ans + w[i] * (val1 + val2) % 1000000007) % 1000000007;
  printf("%lld\n", (ans + 1000000007) % 1000000007);
  return 0;
}

```