#include <bits/stdc++.h>
inline int read() {
  char c = getchar();
  int x = 0;
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c & 15);
    c = getchar();
  }
  return x;
}
const int p = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  return x >= p ? x - p : x;
}
inline int sub(int x, int y) {
  x -= y;
  return x >= 0 ? x : x + p;
}
inline int power(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1ll * x * x % p) {
    if (y & 1) {
      res = 1ll * res * x % p;
    }
  }
  return res;
}
const int N = 200005;
int n, m, sum, ans, f[N], fac[N], ivf[N], ivg[N];
void initCombin(int n) {
  for (int i = fac[0] = 1; i <= n; i++) {
    fac[i] = 1ll * fac[i - 1] * i % p;
  }
  ivf[n] = power(fac[n], p - 2);
  for (int i = n; i; i--) {
    ivf[i - 1] = 1ll * ivf[i] * i % p;
  }
  for (int i = 0; i <= n; i++) {
    ivg[i] = i & 1 ? sub(0, ivf[i]) : ivf[i];
  }
}
int stirling(int k) {
  int res = 0;
  for (int i = 0; i <= k; i++) {
    res = add(res, 1ll * ivg[k - i] * ivf[i] % p * f[i] % p);
  }
  return res;
}
int main() {
  n = read();
  m = read();
  initCombin(m);
  for (int i = 1; i <= n; i++) {
    sum = add(sum, read());
  }
  for (int i = 0; i <= m; i++) {
    f[i] = power(i, n - 1);
  }
  ans = (n - 1ll) * stirling(m) % p;
  for (int i = 0; i <= m; i++) {
    f[i] = 1ll * f[i] * i % p;
  }
  printf("%I64d\n", 1ll * sum * add(ans, stirling(m)) % p);
  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>
inline int read() {
  char c = getchar();
  int x = 0;
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c & 15);
    c = getchar();
  }
  return x;
}
const int p = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  return x >= p ? x - p : x;
}
inline int sub(int x, int y) {
  x -= y;
  return x >= 0 ? x : x + p;
}
inline int power(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1ll * x * x % p) {
    if (y & 1) {
      res = 1ll * res * x % p;
    }
  }
  return res;
}
const int N = 200005;
int n, m, sum, ans, f[N], fac[N], ivf[N], ivg[N];
void initCombin(int n) {
  for (int i = fac[0] = 1; i <= n; i++) {
    fac[i] = 1ll * fac[i - 1] * i % p;
  }
  ivf[n] = power(fac[n], p - 2);
  for (int i = n; i; i--) {
    ivf[i - 1] = 1ll * ivf[i] * i % p;
  }
  for (int i = 0; i <= n; i++) {
    ivg[i] = i & 1 ? sub(0, ivf[i]) : ivf[i];
  }
}
int stirling(int k) {
  int res = 0;
  for (int i = 0; i <= k; i++) {
    res = add(res, 1ll * ivg[k - i] * ivf[i] % p * f[i] % p);
  }
  return res;
}
int main() {
  n = read();
  m = read();
  initCombin(m);
  for (int i = 1; i <= n; i++) {
    sum = add(sum, read());
  }
  for (int i = 0; i <= m; i++) {
    f[i] = power(i, n - 1);
  }
  ans = (n - 1ll) * stirling(m) % p;
  for (int i = 0; i <= m; i++) {
    f[i] = 1ll * f[i] * i % p;
  }
  printf("%I64d\n", 1ll * sum * add(ans, stirling(m)) % p);
  return 0;
}

```