#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000;
const int mod = 1e9 + 7;
const long double Pi = acos(-1.0);
const double eps = 1e-7;
template <class T>
int read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    f |= (ch == '-');
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  x = f ? -x : x;
  return x;
}
int n, k, w, sum, ans, num;
int fac[maxn + 5], invf[maxn + 5];
inline int inc(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int power(int a, int b, int p) {
  int ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) ans = 1ll * ans * a % p;
    a = 1ll * a * a % p;
  }
  return ans;
}
inline int S(int n, int m) {
  int sum = 0;
  for (int i = 0; i <= m; i++) {
    int num = 1ll * power(i, n, mod) * invf[m - i] % mod * invf[i] % mod;
    if ((m - i) & 1) num = mod - num;
    sum = inc(sum, num);
  }
  return sum;
}
int main() {
  read(n);
  read(k);
  fac[0] = invf[0] = 1;
  for (int i = 1; i <= n; i++) sum = inc(sum, read(w));
  int maxx = max(n, k);
  for (int i = 1; i <= maxx; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  invf[maxx] = power(fac[maxx], mod - 2, mod);
  for (int i = maxx - 1; i >= 1; i--)
    invf[i] = 1ll * invf[i + 1] * (i + 1) % mod;
  ans = 1ll * sum * inc(S(n, k), 1ll * (n - 1) * S(n - 1, k) % mod) % mod;
  printf("%d", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 maxn = 200000;
const int mod = 1e9 + 7;
const long double Pi = acos(-1.0);
const double eps = 1e-7;
template <class T>
int read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    f |= (ch == '-');
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  x = f ? -x : x;
  return x;
}
int n, k, w, sum, ans, num;
int fac[maxn + 5], invf[maxn + 5];
inline int inc(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int power(int a, int b, int p) {
  int ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) ans = 1ll * ans * a % p;
    a = 1ll * a * a % p;
  }
  return ans;
}
inline int S(int n, int m) {
  int sum = 0;
  for (int i = 0; i <= m; i++) {
    int num = 1ll * power(i, n, mod) * invf[m - i] % mod * invf[i] % mod;
    if ((m - i) & 1) num = mod - num;
    sum = inc(sum, num);
  }
  return sum;
}
int main() {
  read(n);
  read(k);
  fac[0] = invf[0] = 1;
  for (int i = 1; i <= n; i++) sum = inc(sum, read(w));
  int maxx = max(n, k);
  for (int i = 1; i <= maxx; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  invf[maxx] = power(fac[maxx], mod - 2, mod);
  for (int i = maxx - 1; i >= 1; i--)
    invf[i] = 1ll * invf[i + 1] * (i + 1) % mod;
  ans = 1ll * sum * inc(S(n, k), 1ll * (n - 1) * S(n - 1, k) % mod) % mod;
  printf("%d", ans);
  return 0;
}

```