#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chkmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
  x = x < y ? x : y;
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= f;
}
const int N = 200010, mod = 1e9 + 7;
int main() {
  int n, k;
  read(n), read(k);
  static int fac[N], Inv[N], inv[N];
  fac[0] = fac[1] = Inv[0] = Inv[1] = inv[1] = 1;
  for (int i = 2; i <= n; i++) {
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    Inv[i] = 1ll * Inv[i - 1] * inv[i] % mod;
    fac[i] = 1ll * fac[i - 1] * i % mod;
  }
  auto C = [&](int n, int m) {
    return 1ll * fac[n] * Inv[m] % mod * Inv[n - m] % mod;
  };
  auto Pow = [&](int x, int y) {
    int ret = 1;
    for (; y; y >>= 1, x = 1ll * x * x % mod)
      if (y & 1) ret = 1ll * ret * x % mod;
    return ret;
  };
  auto S = [&](int n, int m) {
    int ret = 0;
    for (int i = 0; i <= m; i++) {
      int ts = 1ll * C(m, i) * Pow(m - i, n) % mod;
      if (i & 1)
        (ret += mod - ts) %= mod;
      else
        (ret += ts) %= mod;
    }
    return 1ll * ret * Inv[m] % mod;
  };
  int sx = (S(n, k) + 1ll * (n - 1) * S(n - 1, k) % mod) % mod, sy = 0;
  for (int i = 1, x; i <= n; i++) cin >> x, (sy += x) %= mod;
  cout << 1ll * sx * sy % mod << '\n';
  return 0;
}

### Prompt

Please formulate a Cpp solution to 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;
template <typename T>
void chkmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
  x = x < y ? x : y;
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= f;
}
const int N = 200010, mod = 1e9 + 7;
int main() {
  int n, k;
  read(n), read(k);
  static int fac[N], Inv[N], inv[N];
  fac[0] = fac[1] = Inv[0] = Inv[1] = inv[1] = 1;
  for (int i = 2; i <= n; i++) {
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    Inv[i] = 1ll * Inv[i - 1] * inv[i] % mod;
    fac[i] = 1ll * fac[i - 1] * i % mod;
  }
  auto C = [&](int n, int m) {
    return 1ll * fac[n] * Inv[m] % mod * Inv[n - m] % mod;
  };
  auto Pow = [&](int x, int y) {
    int ret = 1;
    for (; y; y >>= 1, x = 1ll * x * x % mod)
      if (y & 1) ret = 1ll * ret * x % mod;
    return ret;
  };
  auto S = [&](int n, int m) {
    int ret = 0;
    for (int i = 0; i <= m; i++) {
      int ts = 1ll * C(m, i) * Pow(m - i, n) % mod;
      if (i & 1)
        (ret += mod - ts) %= mod;
      else
        (ret += ts) %= mod;
    }
    return 1ll * ret * Inv[m] % mod;
  };
  int sx = (S(n, k) + 1ll * (n - 1) * S(n - 1, k) % mod) % mod, sy = 0;
  for (int i = 1, x; i <= n; i++) cin >> x, (sy += x) %= mod;
  cout << 1ll * sx * sy % mod << '\n';
  return 0;
}

```