#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long sum = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = sum * 10 + ch - '0';
    ch = getchar();
  }
  return f * sum;
}
const long long N = 200001;
const long long Mod = 1000000007;
long long n, m, sum, f, jc[N];
inline long long fp(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (long long)ret * a % Mod;
    a = (long long)a * a % Mod;
    b >>= 1;
  }
  return ret;
}
inline long long inv(long long x) { return fp(x, Mod - 2); }
inline long long stir(long long n, long long k) {
  long long ret = 0;
  for (long long j = 0; j < k; j++)
    ret = (ret + fp(k - j, n) * inv(jc[j]) % Mod * inv(jc[k - j]) *
                     ((j & 1) ? -1 : 1)) %
          Mod;
  return ret;
}
signed main() {
  n = read();
  m = read();
  jc[0] = 1;
  for (long long i = 1; i <= n; i++) sum = (sum + read()) % Mod;
  for (long long i = 1; i <= m; i++) jc[i] = jc[i - 1] * i % Mod;
  long long ans = sum * (stir(n, m) + stir(n - 1, m) * (n - 1) % Mod) % Mod;
  printf("%lld\n", (ans + Mod) % Mod);
  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;
inline long long read() {
  long long sum = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = sum * 10 + ch - '0';
    ch = getchar();
  }
  return f * sum;
}
const long long N = 200001;
const long long Mod = 1000000007;
long long n, m, sum, f, jc[N];
inline long long fp(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (long long)ret * a % Mod;
    a = (long long)a * a % Mod;
    b >>= 1;
  }
  return ret;
}
inline long long inv(long long x) { return fp(x, Mod - 2); }
inline long long stir(long long n, long long k) {
  long long ret = 0;
  for (long long j = 0; j < k; j++)
    ret = (ret + fp(k - j, n) * inv(jc[j]) % Mod * inv(jc[k - j]) *
                     ((j & 1) ? -1 : 1)) %
          Mod;
  return ret;
}
signed main() {
  n = read();
  m = read();
  jc[0] = 1;
  for (long long i = 1; i <= n; i++) sum = (sum + read()) % Mod;
  for (long long i = 1; i <= m; i++) jc[i] = jc[i - 1] * i % Mod;
  long long ans = sum * (stir(n, m) + stir(n - 1, m) * (n - 1) % Mod) % Mod;
  printf("%lld\n", (ans + Mod) % Mod);
  return 0;
}

```