#include <bits/stdc++.h>
using namespace std;
const int SIZE = 1 << 10;
int pointer = SIZE;
char buffer[SIZE];
char Advance() {
  if (pointer == SIZE) {
    fread(buffer, 1, SIZE, stdin);
    pointer = 0;
  }
  return buffer[pointer++];
}
int Read() {
  int answer = 0;
  char ch = Advance();
  while (!isdigit(ch)) ch = Advance();
  while (isdigit(ch)) {
    answer = answer * 10 + ch - '0';
    ch = Advance();
  }
  return answer;
}
const int MAXN = 200000;
const int MOD = 1000000007;
int v[1 + MAXN];
void Add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
void Subtract(int &x, int y) {
  x -= y;
  if (x < 0) x += MOD;
}
int factorial[1 + MAXN], inverse[1 + MAXN];
int RaiseToPower(int base, int power) {
  int answer = 1;
  while (power) {
    if (power % 2) answer = (1LL * answer * base) % MOD;
    base = (1LL * base * base) % MOD;
    power /= 2;
  }
  return answer;
}
void Precompute(int n) {
  factorial[0] = inverse[0] = 1;
  for (int i = 1; i <= n; i++)
    factorial[i] = (1LL * factorial[i - 1] * i) % MOD;
  inverse[n] = RaiseToPower(factorial[n], MOD - 2);
  for (int i = n - 1; i >= 1; i--)
    inverse[i] = (1LL * (i + 1) * inverse[i + 1]) % MOD;
}
int Combinations(int n, int k) {
  int answer = (1LL * inverse[n - k] * inverse[k]) % MOD;
  answer = (1LL * answer * factorial[n]) % MOD;
  return answer;
}
int Stirling(int n, int k) {
  int answer = 0, sign = 1;
  for (int i = k; i >= 0; i--) {
    int add = (1LL * Combinations(k, i) * RaiseToPower(i, n)) % MOD;
    if (sign)
      Add(answer, add);
    else
      Subtract(answer, add);
    sign = 1 - sign;
  }
  answer = (1LL * answer * inverse[k]) % MOD;
  return answer;
}
int main() {
  int n, k, sum = 0;
  scanf("%d%d", &n, &k);
  Precompute(n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    Add(sum, x);
  }
  if (k == n) {
    printf("%d\n", sum);
    return 0;
  }
  if (k == 1) {
    sum = (1LL * sum * n) % MOD;
    printf("%d\n", sum);
    return 0;
  }
  int first = Stirling(n, k),
      second = (1LL * (n - 1) * Stirling(n - 1, k)) % MOD;
  sum = (1LL * sum * (first + second)) % MOD;
  printf("%d\n", sum);
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
const int SIZE = 1 << 10;
int pointer = SIZE;
char buffer[SIZE];
char Advance() {
  if (pointer == SIZE) {
    fread(buffer, 1, SIZE, stdin);
    pointer = 0;
  }
  return buffer[pointer++];
}
int Read() {
  int answer = 0;
  char ch = Advance();
  while (!isdigit(ch)) ch = Advance();
  while (isdigit(ch)) {
    answer = answer * 10 + ch - '0';
    ch = Advance();
  }
  return answer;
}
const int MAXN = 200000;
const int MOD = 1000000007;
int v[1 + MAXN];
void Add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
void Subtract(int &x, int y) {
  x -= y;
  if (x < 0) x += MOD;
}
int factorial[1 + MAXN], inverse[1 + MAXN];
int RaiseToPower(int base, int power) {
  int answer = 1;
  while (power) {
    if (power % 2) answer = (1LL * answer * base) % MOD;
    base = (1LL * base * base) % MOD;
    power /= 2;
  }
  return answer;
}
void Precompute(int n) {
  factorial[0] = inverse[0] = 1;
  for (int i = 1; i <= n; i++)
    factorial[i] = (1LL * factorial[i - 1] * i) % MOD;
  inverse[n] = RaiseToPower(factorial[n], MOD - 2);
  for (int i = n - 1; i >= 1; i--)
    inverse[i] = (1LL * (i + 1) * inverse[i + 1]) % MOD;
}
int Combinations(int n, int k) {
  int answer = (1LL * inverse[n - k] * inverse[k]) % MOD;
  answer = (1LL * answer * factorial[n]) % MOD;
  return answer;
}
int Stirling(int n, int k) {
  int answer = 0, sign = 1;
  for (int i = k; i >= 0; i--) {
    int add = (1LL * Combinations(k, i) * RaiseToPower(i, n)) % MOD;
    if (sign)
      Add(answer, add);
    else
      Subtract(answer, add);
    sign = 1 - sign;
  }
  answer = (1LL * answer * inverse[k]) % MOD;
  return answer;
}
int main() {
  int n, k, sum = 0;
  scanf("%d%d", &n, &k);
  Precompute(n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    Add(sum, x);
  }
  if (k == n) {
    printf("%d\n", sum);
    return 0;
  }
  if (k == 1) {
    sum = (1LL * sum * n) % MOD;
    printf("%d\n", sum);
    return 0;
  }
  int first = Stirling(n, k),
      second = (1LL * (n - 1) * Stirling(n - 1, k)) % MOD;
  sum = (1LL * sum * (first + second)) % MOD;
  printf("%d\n", sum);
  return 0;
}

```