#include <bits/stdc++.h>
using namespace std;
void qmax(int &x, int y) {
  if (x < y) x = y;
}
void qmin(int &x, int y) {
  if (x > y) x = y;
}
inline int read() {
  char s;
  int k = 0, base = 1;
  while ((s = getchar()) != '-' && s != EOF && !(isdigit(s)))
    ;
  if (s == EOF) exit(0);
  if (s == '-') base = -1, s = getchar();
  while (isdigit(s)) {
    k = k * 10 + (s ^ '0');
    s = getchar();
  }
  return k * base;
}
inline void write(int x) {
  static char cnt, num[15];
  cnt = 0;
  if (!x) {
    printf("0");
    return;
  }
  for (; x; x /= 10) num[++cnt] = x % 10;
  for (; cnt; putchar(num[cnt--] + 48))
    ;
}
const long long mod = 1e9 + 7;
long long ksm(long long x, int y) {
  long long sum = 1;
  while (y) {
    if (y & 1) sum = sum * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return sum;
}
const int maxn = 2e5 + 100;
long long n, k, ans, S, jie[maxn], ny[maxn], X, S1;
int main() {
  n = read();
  k = read();
  jie[0] = 1;
  for (int i = 1; i <= k; i++) jie[i] = jie[i - 1] * i % mod;
  ny[k] = ksm(jie[k], mod - 2);
  for (int i = k - 1; i >= 0; i--) ny[i] = ny[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= k; i++) {
    if ((k - i) % 2 == 1) {
      S = S - jie[k] * ny[i] % mod * ny[k - i] % mod * ksm(i, n) % mod;
      if (S < 0) S += mod;
    } else {
      S = S + jie[k] * ny[i] % mod * ny[k - i] % mod * ksm(i, n) % mod;
      if (S >= mod) S -= mod;
    }
  }
  S = S * ny[k] % mod;
  S1 = S;
  S = 0;
  n--;
  for (int i = 1; i <= k; i++) {
    if ((k - i) % 2 == 1) {
      S = S - jie[k] * ny[i] % mod * ny[k - i] % mod * ksm(i, n) % mod;
      if (S < 0) S += mod;
    } else {
      S = S + jie[k] * ny[i] % mod * ny[k - i] % mod * ksm(i, n) % mod;
      if (S >= mod) S -= mod;
    }
  }
  S = S * ny[k] % mod;
  S = n * S % mod + S1;
  S %= mod;
  n++;
  for (int i = 1; i <= n; i++) {
    X = read();
    ans = ans + S * X % mod;
    if (ans >= mod) ans -= mod;
  }
  printf("%lld\n", ans);
  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;
void qmax(int &x, int y) {
  if (x < y) x = y;
}
void qmin(int &x, int y) {
  if (x > y) x = y;
}
inline int read() {
  char s;
  int k = 0, base = 1;
  while ((s = getchar()) != '-' && s != EOF && !(isdigit(s)))
    ;
  if (s == EOF) exit(0);
  if (s == '-') base = -1, s = getchar();
  while (isdigit(s)) {
    k = k * 10 + (s ^ '0');
    s = getchar();
  }
  return k * base;
}
inline void write(int x) {
  static char cnt, num[15];
  cnt = 0;
  if (!x) {
    printf("0");
    return;
  }
  for (; x; x /= 10) num[++cnt] = x % 10;
  for (; cnt; putchar(num[cnt--] + 48))
    ;
}
const long long mod = 1e9 + 7;
long long ksm(long long x, int y) {
  long long sum = 1;
  while (y) {
    if (y & 1) sum = sum * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return sum;
}
const int maxn = 2e5 + 100;
long long n, k, ans, S, jie[maxn], ny[maxn], X, S1;
int main() {
  n = read();
  k = read();
  jie[0] = 1;
  for (int i = 1; i <= k; i++) jie[i] = jie[i - 1] * i % mod;
  ny[k] = ksm(jie[k], mod - 2);
  for (int i = k - 1; i >= 0; i--) ny[i] = ny[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= k; i++) {
    if ((k - i) % 2 == 1) {
      S = S - jie[k] * ny[i] % mod * ny[k - i] % mod * ksm(i, n) % mod;
      if (S < 0) S += mod;
    } else {
      S = S + jie[k] * ny[i] % mod * ny[k - i] % mod * ksm(i, n) % mod;
      if (S >= mod) S -= mod;
    }
  }
  S = S * ny[k] % mod;
  S1 = S;
  S = 0;
  n--;
  for (int i = 1; i <= k; i++) {
    if ((k - i) % 2 == 1) {
      S = S - jie[k] * ny[i] % mod * ny[k - i] % mod * ksm(i, n) % mod;
      if (S < 0) S += mod;
    } else {
      S = S + jie[k] * ny[i] % mod * ny[k - i] % mod * ksm(i, n) % mod;
      if (S >= mod) S -= mod;
    }
  }
  S = S * ny[k] % mod;
  S = n * S % mod + S1;
  S %= mod;
  n++;
  for (int i = 1; i <= n; i++) {
    X = read();
    ans = ans + S * X % mod;
    if (ans >= mod) ans -= mod;
  }
  printf("%lld\n", ans);
  return 0;
}

```