#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
inline int add(int a, int b) {
  if ((a += b) >= mod) a -= mod;
  return a;
}
inline int mult(int a, int b) {
  long long t = 1ll * a * b;
  if (t >= mod) t %= mod;
  return t;
}
inline int power(int a, int b) {
  int out = 1;
  while (b) {
    if (b & 1) out = mult(out, a);
    a = mult(a, a);
    b >>= 1;
  }
  return out;
}
int n, K, w[200010], fac[200010], ifac[200010];
inline int C(int a, int b) {
  if (a < b) return 0;
  return mult(fac[a], mult(ifac[b], ifac[a - b]));
}
inline int S(int n, int k) {
  int ret = 0;
  for (int i = 0; i <= k; i++)
    ret = add(ret, mult(power(mod - 1, i), mult(C(k, i), power(k - i, n))));
  return mult(ret, ifac[k]);
}
int main() {
  scanf("%d%d", &n, &K);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = mult(fac[i - 1], i);
  ifac[n] = power(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) ifac[i] = mult(ifac[i + 1], i + 1);
  int sum = 0;
  for (int i = 1, ti; i <= n; i++) scanf("%d", &ti), sum = add(sum, ti);
  printf("%d\n", mult(add(S(n, K), mult(n - 1, S(n - 1, K))), sum));
  return 0;
}

### Prompt

Create a solution in Cpp for 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 mod = 1000000007;
inline int add(int a, int b) {
  if ((a += b) >= mod) a -= mod;
  return a;
}
inline int mult(int a, int b) {
  long long t = 1ll * a * b;
  if (t >= mod) t %= mod;
  return t;
}
inline int power(int a, int b) {
  int out = 1;
  while (b) {
    if (b & 1) out = mult(out, a);
    a = mult(a, a);
    b >>= 1;
  }
  return out;
}
int n, K, w[200010], fac[200010], ifac[200010];
inline int C(int a, int b) {
  if (a < b) return 0;
  return mult(fac[a], mult(ifac[b], ifac[a - b]));
}
inline int S(int n, int k) {
  int ret = 0;
  for (int i = 0; i <= k; i++)
    ret = add(ret, mult(power(mod - 1, i), mult(C(k, i), power(k - i, n))));
  return mult(ret, ifac[k]);
}
int main() {
  scanf("%d%d", &n, &K);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = mult(fac[i - 1], i);
  ifac[n] = power(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) ifac[i] = mult(ifac[i + 1], i + 1);
  int sum = 0;
  for (int i = 1, ti; i <= n; i++) scanf("%d", &ti), sum = add(sum, ti);
  printf("%d\n", mult(add(S(n, K), mult(n - 1, S(n - 1, K))), sum));
  return 0;
}

```