#include <bits/stdc++.h>
using namespace std;
const int CN = 2e5 + 10;
const int P = 1e9 + 7;
int read() {
  int s = 0, ne = 1;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') ne = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) s = (s << 1) + (s << 3) + c - '0';
  return s * ne;
}
int add(int x, int y) { return x + y >= P ? x + y - P : x + y; }
int qp(int a, int b) {
  int r = 1;
  for (; b; a = 1ll * a * a % P, b >>= 1)
    if (b & 1) r = 1ll * r * a % P;
  return r;
}
int fac[CN], iv[CN], ifac[CN];
void init(int n) {
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * i * fac[i - 1] % P;
  iv[1] = 1;
  for (int i = 2; i <= n; i++) iv[i] = 1ll * iv[P % i] * (P - P / i) % P;
  ifac[0] = 1;
  for (int i = 1; i <= n; i++) ifac[i] = 1ll * ifac[i - 1] * iv[i] % P;
}
int C(int n, int m) { return 1ll * fac[n] * ifac[m] % P * ifac[n - m] % P; }
int S(int n, int m) {
  int res = 0;
  for (int i = 0; i <= m; i++) {
    int cur = 1ll * C(m, i) * qp(m - i, n) % P;
    res = add(res, i & 1 ? P - cur : cur);
  }
  return 1ll * res * ifac[m] % P;
}
int n, k, s, ans;
int main() {
  init(2e5), n = read(), k = read();
  for (int i = 0; i < n; i++) s = add(s, read());
  return ans = 1ll * s * add(S(n, k), 1ll * (n - 1) * S(n - 1, k) % P) % P,
         printf("%d\n", ans), 0;
}

### Prompt

In Cpp, your task is to solve 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;
const int CN = 2e5 + 10;
const int P = 1e9 + 7;
int read() {
  int s = 0, ne = 1;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') ne = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) s = (s << 1) + (s << 3) + c - '0';
  return s * ne;
}
int add(int x, int y) { return x + y >= P ? x + y - P : x + y; }
int qp(int a, int b) {
  int r = 1;
  for (; b; a = 1ll * a * a % P, b >>= 1)
    if (b & 1) r = 1ll * r * a % P;
  return r;
}
int fac[CN], iv[CN], ifac[CN];
void init(int n) {
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * i * fac[i - 1] % P;
  iv[1] = 1;
  for (int i = 2; i <= n; i++) iv[i] = 1ll * iv[P % i] * (P - P / i) % P;
  ifac[0] = 1;
  for (int i = 1; i <= n; i++) ifac[i] = 1ll * ifac[i - 1] * iv[i] % P;
}
int C(int n, int m) { return 1ll * fac[n] * ifac[m] % P * ifac[n - m] % P; }
int S(int n, int m) {
  int res = 0;
  for (int i = 0; i <= m; i++) {
    int cur = 1ll * C(m, i) * qp(m - i, n) % P;
    res = add(res, i & 1 ? P - cur : cur);
  }
  return 1ll * res * ifac[m] % P;
}
int n, k, s, ans;
int main() {
  init(2e5), n = read(), k = read();
  for (int i = 0; i < n; i++) s = add(s, read());
  return ans = 1ll * s * add(S(n, k), 1ll * (n - 1) * S(n - 1, k) % P) % P,
         printf("%d\n", ans), 0;
}

```