#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool ckmin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline bool ckmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int N = 2e5 + 10, P = 1e9 + 7;
int fac[N], ifac[N], inv[N];
int add(int first) { return first >= P ? first - P : first; }
int sub(int first) { return first < 0 ? first + P : first; }
void Add(int &first, int second) {
  if ((first += second) >= P && (first -= P))
    ;
}
void Sub(int &first, int second) {
  if ((first -= second) < 0 && (first += P))
    ;
}
int qpow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % P)
    if (b & 1) res = (long long)res * a % P;
  return res;
}
void init(int n) {
  inv[1] = fac[0] = ifac[0] = 1;
  for (int i = 2; i <= n; i++) inv[i] = (long long)(P - P / i) * inv[P % i] % P;
  for (int i = 1; i <= n; i++) {
    fac[i] = 1ll * fac[i - 1] * i % P;
    ifac[i] = 1ll * ifac[i - 1] * inv[i] % P;
  }
  return;
}
int S2(int n, int m) {
  int res = 0;
  for (int i = 0; i <= m; i++) {
    int tmp = 1ll * fac[m] * ifac[i] % P * ifac[m - i] % P * qpow(m - i, n) % P;
    if (i & 1)
      Sub(res, tmp);
    else
      Add(res, tmp);
  }
  return 1ll * res * ifac[m] % P;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  init(n);
  int s = 0;
  for (int i = 1; i <= n; i++) {
    int first;
    scanf("%d", &first);
    s = (s + first) % P;
  }
  int ans = add(1ll * (n - 1) * S2(n - 1, k) % P + S2(n, k));
  ans = 1ll * ans * s % P;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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;
template <typename T>
inline bool ckmin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline bool ckmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int N = 2e5 + 10, P = 1e9 + 7;
int fac[N], ifac[N], inv[N];
int add(int first) { return first >= P ? first - P : first; }
int sub(int first) { return first < 0 ? first + P : first; }
void Add(int &first, int second) {
  if ((first += second) >= P && (first -= P))
    ;
}
void Sub(int &first, int second) {
  if ((first -= second) < 0 && (first += P))
    ;
}
int qpow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % P)
    if (b & 1) res = (long long)res * a % P;
  return res;
}
void init(int n) {
  inv[1] = fac[0] = ifac[0] = 1;
  for (int i = 2; i <= n; i++) inv[i] = (long long)(P - P / i) * inv[P % i] % P;
  for (int i = 1; i <= n; i++) {
    fac[i] = 1ll * fac[i - 1] * i % P;
    ifac[i] = 1ll * ifac[i - 1] * inv[i] % P;
  }
  return;
}
int S2(int n, int m) {
  int res = 0;
  for (int i = 0; i <= m; i++) {
    int tmp = 1ll * fac[m] * ifac[i] % P * ifac[m - i] % P * qpow(m - i, n) % P;
    if (i & 1)
      Sub(res, tmp);
    else
      Add(res, tmp);
  }
  return 1ll * res * ifac[m] % P;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  init(n);
  int s = 0;
  for (int i = 1; i <= n; i++) {
    int first;
    scanf("%d", &first);
    s = (s + first) % P;
  }
  int ans = add(1ll * (n - 1) * S2(n - 1, k) % P + S2(n, k));
  ans = 1ll * ans * s % P;
  printf("%d\n", ans);
  return 0;
}

```