#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 1;
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= neg;
}
const int MAXK = 2e5;
const int MOD = 1e9 + 7;
int qpow(int x, int e) {
  if (e < 0) return 0;
  int ret = 1;
  for (; e; e >>= 1, x = 1ll * x * x % MOD)
    if (e & 1) ret = 1ll * ret * x % MOD;
  return ret;
}
int fac[MAXK + 5], ifac[MAXK + 5];
void init_fac(int n) {
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % MOD;
  ifac[n] = qpow(fac[n], MOD - 2);
  for (int i = n - 1; ~i; i--) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % MOD;
}
int n, k, sum = 0, ans = 0;
int main() {
  scanf("%d%d", &n, &k);
  init_fac(k);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    (sum += x) %= MOD;
  }
  for (int i = 0; i < k; i++) {
    if ((k - 1 - i) & 1)
      ans =
          (ans -
           1ll * ifac[i] % MOD * ifac[k - 1 - i] % MOD *
               (qpow(i + 1, n - 1) + 1ll * (n - 1) * qpow(i + 1, n - 2) % MOD) %
               MOD +
           MOD) %
          MOD;
    else
      ans = (ans + 1ll * ifac[i] % MOD * ifac[k - 1 - i] % MOD *
                       (qpow(i + 1, n - 1) +
                        1ll * (n - 1) * qpow(i + 1, n - 2) % MOD) %
                       MOD) %
            MOD;
  }
  printf("%d\n", 1ll * sum * ans % MOD);
  return 0;
}

### Prompt

Your task is to create 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 T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 1;
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= neg;
}
const int MAXK = 2e5;
const int MOD = 1e9 + 7;
int qpow(int x, int e) {
  if (e < 0) return 0;
  int ret = 1;
  for (; e; e >>= 1, x = 1ll * x * x % MOD)
    if (e & 1) ret = 1ll * ret * x % MOD;
  return ret;
}
int fac[MAXK + 5], ifac[MAXK + 5];
void init_fac(int n) {
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % MOD;
  ifac[n] = qpow(fac[n], MOD - 2);
  for (int i = n - 1; ~i; i--) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % MOD;
}
int n, k, sum = 0, ans = 0;
int main() {
  scanf("%d%d", &n, &k);
  init_fac(k);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    (sum += x) %= MOD;
  }
  for (int i = 0; i < k; i++) {
    if ((k - 1 - i) & 1)
      ans =
          (ans -
           1ll * ifac[i] % MOD * ifac[k - 1 - i] % MOD *
               (qpow(i + 1, n - 1) + 1ll * (n - 1) * qpow(i + 1, n - 2) % MOD) %
               MOD +
           MOD) %
          MOD;
    else
      ans = (ans + 1ll * ifac[i] % MOD * ifac[k - 1 - i] % MOD *
                       (qpow(i + 1, n - 1) +
                        1ll * (n - 1) * qpow(i + 1, n - 2) % MOD) %
                       MOD) %
            MOD;
  }
  printf("%d\n", 1ll * sum * ans % MOD);
  return 0;
}

```