#include <bits/stdc++.h>
using namespace std;
struct pat {
  int x, y;
  pat(int x = 0, int y = 0) : x(x), y(y) {}
  bool operator<(const pat &p) const { return x == p.x ? y < p.y : x < p.x; }
};
template <typename TP>
inline void read(TP &tar) {
  TP ret = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + (ch - '0');
    ch = getchar();
  }
  tar = ret * f;
}
namespace RKK {
const int N = 200011, mo = 1000000007;
void doadd(long long &a, long long b) {
  if ((a += b) >= mo) a -= mo;
}
long long fpow(long long a, long long p) {
  long long ret = 1;
  while (p) {
    if (p & 1ll) (ret *= a) %= mo;
    (a *= a) %= mo, p >>= 1;
  }
  return ret;
}
long long fac[N], inv[N], ifac[N];
void init() {
  inv[1] = 1;
  for (int i = 2; i <= 200000; i++) inv[i] = inv[mo % i] * (mo - mo / i) % mo;
  fac[0] = 1;
  for (int i = 1; i <= 200000; i++) fac[i] = fac[i - 1] * i % mo;
  ifac[0] = 1;
  for (int i = 1; i <= 200000; i++) ifac[i] = ifac[i - 1] * inv[i] % mo;
}
int n, k;
long long xi;
long long w, ans;
int main() {
  init(), read(n), read(k);
  for (int i = 1; i <= n; i++) read(xi), doadd(w, xi);
  for (int i = 0; i <= k; i++)
    doadd(ans, fpow(k - i, n - 1) * ifac[i] % mo * ifac[k - i] % mo *
                   ((i & 1) ? mo - 1 : 1) % mo);
  (ans *= (n - 1)) %= mo;
  for (int i = 0; i <= k; i++)
    doadd(ans, fpow(k - i, n) * ifac[i] % mo * ifac[k - i] % mo *
                   ((i & 1) ? mo - 1 : 1) % mo);
  printf("%I64d\n", ans * w % mo);
  return 0;
}
}  // namespace RKK
int main() { return RKK::main(); }

### 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;
struct pat {
  int x, y;
  pat(int x = 0, int y = 0) : x(x), y(y) {}
  bool operator<(const pat &p) const { return x == p.x ? y < p.y : x < p.x; }
};
template <typename TP>
inline void read(TP &tar) {
  TP ret = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + (ch - '0');
    ch = getchar();
  }
  tar = ret * f;
}
namespace RKK {
const int N = 200011, mo = 1000000007;
void doadd(long long &a, long long b) {
  if ((a += b) >= mo) a -= mo;
}
long long fpow(long long a, long long p) {
  long long ret = 1;
  while (p) {
    if (p & 1ll) (ret *= a) %= mo;
    (a *= a) %= mo, p >>= 1;
  }
  return ret;
}
long long fac[N], inv[N], ifac[N];
void init() {
  inv[1] = 1;
  for (int i = 2; i <= 200000; i++) inv[i] = inv[mo % i] * (mo - mo / i) % mo;
  fac[0] = 1;
  for (int i = 1; i <= 200000; i++) fac[i] = fac[i - 1] * i % mo;
  ifac[0] = 1;
  for (int i = 1; i <= 200000; i++) ifac[i] = ifac[i - 1] * inv[i] % mo;
}
int n, k;
long long xi;
long long w, ans;
int main() {
  init(), read(n), read(k);
  for (int i = 1; i <= n; i++) read(xi), doadd(w, xi);
  for (int i = 0; i <= k; i++)
    doadd(ans, fpow(k - i, n - 1) * ifac[i] % mo * ifac[k - i] % mo *
                   ((i & 1) ? mo - 1 : 1) % mo);
  (ans *= (n - 1)) %= mo;
  for (int i = 0; i <= k; i++)
    doadd(ans, fpow(k - i, n) * ifac[i] % mo * ifac[k - i] % mo *
                   ((i & 1) ? mo - 1 : 1) % mo);
  printf("%I64d\n", ans * w % mo);
  return 0;
}
}  // namespace RKK
int main() { return RKK::main(); }

```