#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void read(T &x) {
  x = 0;
  register char c = getchar();
  bool f = 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') c = getchar(), f = 1;
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  x = f ? -x : x;
  return;
}
template <class T>
inline void write(T x) {
  if (!x) {
    putchar(48);
    return;
  }
  if (x < 0) x = -x, putchar('-');
  int len = -1, z[20];
  while (x > 0) z[++len] = x % 10, x /= 10;
  for (register int i = len; i >= 0; i--) putchar(z[i] + 48);
  return;
}
const int N = 200010, Mod = 1e9 + 7;
int fac[N], ifac[N];
int fpow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % Mod)
    if (b & 1) res = 1ll * res * a % Mod;
  return res;
}
int S(int n, int m) {
  int res = 0;
  for (int k = 0; k <= m; k++) {
    int p = 1ll * fpow(m - k, n) * ifac[m - k] % Mod * ifac[k] % Mod;
    if (k & 1)
      res = (res - p + Mod) % Mod;
    else
      res = (res + p) % Mod;
  }
  return res;
}
int main() {
  int n, k, sum = 0;
  read(n), read(k);
  for (int i = 1; i <= n; i++) {
    int x;
    read(x);
    sum = (sum + x) % Mod;
  }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
  ifac[n] = fpow(fac[n], Mod - 2);
  for (int i = n; i >= 1; i--) ifac[i - 1] = 1ll * ifac[i] * i % Mod;
  printf("%d\n",
         (S(n, k) + 1ll * (n - 1) * S(n - 1, k) % Mod) % Mod * sum % Mod);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 <class T>
inline void read(T &x) {
  x = 0;
  register char c = getchar();
  bool f = 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') c = getchar(), f = 1;
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  x = f ? -x : x;
  return;
}
template <class T>
inline void write(T x) {
  if (!x) {
    putchar(48);
    return;
  }
  if (x < 0) x = -x, putchar('-');
  int len = -1, z[20];
  while (x > 0) z[++len] = x % 10, x /= 10;
  for (register int i = len; i >= 0; i--) putchar(z[i] + 48);
  return;
}
const int N = 200010, Mod = 1e9 + 7;
int fac[N], ifac[N];
int fpow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % Mod)
    if (b & 1) res = 1ll * res * a % Mod;
  return res;
}
int S(int n, int m) {
  int res = 0;
  for (int k = 0; k <= m; k++) {
    int p = 1ll * fpow(m - k, n) * ifac[m - k] % Mod * ifac[k] % Mod;
    if (k & 1)
      res = (res - p + Mod) % Mod;
    else
      res = (res + p) % Mod;
  }
  return res;
}
int main() {
  int n, k, sum = 0;
  read(n), read(k);
  for (int i = 1; i <= n; i++) {
    int x;
    read(x);
    sum = (sum + x) % Mod;
  }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
  ifac[n] = fpow(fac[n], Mod - 2);
  for (int i = n; i >= 1; i--) ifac[i - 1] = 1ll * ifac[i] * i % Mod;
  printf("%d\n",
         (S(n, k) + 1ll * (n - 1) * S(n - 1, k) % Mod) % Mod * sum % Mod);
  return 0;
}

```