#include <bits/stdc++.h>
using namespace std;
void exgcd(int a, int b, int& d, int& x, int& y) {
  if (!b)
    d = a, x = 1, y = 0;
  else {
    exgcd(b, a % b, d, y, x);
    y -= (a / b) * x;
  }
}
int inv(int a, int n) {
  int d, x, y;
  exgcd(a, n, d, x, y);
  return (x < 0) ? (x + n) : (x);
}
int qpow(int a, int pos, int m) {
  int rt = 1, pa = a;
  for (; pos; pos >>= 1, pa = pa * 1ll * pa % m)
    if (pos & 1) rt = rt * 1ll * pa % m;
  return rt;
}
const int M = 1e9 + 7;
int n, k;
int sum = 0;
inline void init() {
  scanf("%d%d", &n, &k);
  for (int i = 1, x; i <= n; i++) scanf("%d", &x), sum = (sum + x) % M;
}
inline int Striling(int n, int k) {
  int rt = 0, sign = 1, C = 1, P = 1;
  for (int i = 0; i < k; i++, sign = -sign) {
    if (i) C = (C * 1ll * inv(i, M) % M) * (k - i + 1) % M;
    rt = (rt + sign * ((qpow(k - i, n, M) * 1ll * C) % M) + M) % M;
  }
  for (int i = 2; i <= k; i++) P = (P * 1ll * i) % M;
  return rt * 1ll * inv(P, M) % M;
}
int res = 0;
inline void solve() {
  res = (Striling(n, k) + (n - 1) * 1ll * Striling(n - 1, k)) % M;
  res = res * 1ll * sum % M;
  printf("%d", res);
}
int main() {
  init();
  solve();
  return 0;
}

### Prompt

Please create a solution in CPP 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;
void exgcd(int a, int b, int& d, int& x, int& y) {
  if (!b)
    d = a, x = 1, y = 0;
  else {
    exgcd(b, a % b, d, y, x);
    y -= (a / b) * x;
  }
}
int inv(int a, int n) {
  int d, x, y;
  exgcd(a, n, d, x, y);
  return (x < 0) ? (x + n) : (x);
}
int qpow(int a, int pos, int m) {
  int rt = 1, pa = a;
  for (; pos; pos >>= 1, pa = pa * 1ll * pa % m)
    if (pos & 1) rt = rt * 1ll * pa % m;
  return rt;
}
const int M = 1e9 + 7;
int n, k;
int sum = 0;
inline void init() {
  scanf("%d%d", &n, &k);
  for (int i = 1, x; i <= n; i++) scanf("%d", &x), sum = (sum + x) % M;
}
inline int Striling(int n, int k) {
  int rt = 0, sign = 1, C = 1, P = 1;
  for (int i = 0; i < k; i++, sign = -sign) {
    if (i) C = (C * 1ll * inv(i, M) % M) * (k - i + 1) % M;
    rt = (rt + sign * ((qpow(k - i, n, M) * 1ll * C) % M) + M) % M;
  }
  for (int i = 2; i <= k; i++) P = (P * 1ll * i) % M;
  return rt * 1ll * inv(P, M) % M;
}
int res = 0;
inline void solve() {
  res = (Striling(n, k) + (n - 1) * 1ll * Striling(n - 1, k)) % M;
  res = res * 1ll * sum % M;
  printf("%d", res);
}
int main() {
  init();
  solve();
  return 0;
}

```