#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = x * 10 + (c & 15), c = getchar();
}
const int P = 1000000007;
inline void add(int &a, int b) {
  a += b;
  if (a >= P) a -= P;
}
inline int ksm(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = (long long)a * a % P)
    if (b & 1) s = (long long)s * a % P;
  return s;
}
int n, k, sum, jc[200001], invjc[200001], mi[200001], ans, prime[200001], cnt;
bool vis[200001];
inline int C(int n, int m) {
  return (long long)jc[n] * invjc[m] % P * invjc[n - m] % P;
}
inline int S() {
  int ans = 0;
  for (int i = 0; i <= k; i++)
    add(ans, (long long)((i & 1) ? P - 1 : 1) * C(k, i) % P * mi[k - i] % P);
  return (long long)ans * invjc[k] % P;
}
signed main() {
  read(n), read(k);
  for (int i = 1, x; i <= n; i++) read(x), add(sum, x);
  if (n == k) {
    printf("%d", sum);
    return 0;
  }
  jc[0] = 1;
  for (int i = 1; i <= k; i++) jc[i] = (long long)jc[i - 1] * i % P;
  invjc[k] = ksm(jc[k], P - 2);
  for (int i = k; i >= 1; i--) invjc[i - 1] = (long long)invjc[i] * i % P;
  mi[1] = 1;
  for (int i = 2; i <= k; i++) {
    if (!vis[i]) prime[++cnt] = i, mi[i] = ksm(i, n);
    for (int j = 1; j <= cnt && prime[j] * i <= k; i++) {
      vis[prime[j] * i] = 1;
      mi[prime[j] * i] = (long long)mi[i] * mi[prime[j]] % P;
      if (i % prime[j] == 0) break;
    }
  }
  ans = S();
  for (int i = 2; i <= k; i++) {
    if (!vis[i]) mi[i] = ksm(i, n - 1);
    for (int j = 1; j <= cnt && prime[j] * i <= k; i++) {
      mi[prime[j] * i] = (long long)mi[i] * mi[prime[j]] % P;
      if (i % prime[j] == 0) break;
    }
  }
  add(ans, (long long)(n - 1) * S() % P);
  ans = (long long)ans * sum % P;
  printf("%d", ans);
  return 0;
}

### Prompt

Create a solution in CPP for 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 <class T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = x * 10 + (c & 15), c = getchar();
}
const int P = 1000000007;
inline void add(int &a, int b) {
  a += b;
  if (a >= P) a -= P;
}
inline int ksm(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = (long long)a * a % P)
    if (b & 1) s = (long long)s * a % P;
  return s;
}
int n, k, sum, jc[200001], invjc[200001], mi[200001], ans, prime[200001], cnt;
bool vis[200001];
inline int C(int n, int m) {
  return (long long)jc[n] * invjc[m] % P * invjc[n - m] % P;
}
inline int S() {
  int ans = 0;
  for (int i = 0; i <= k; i++)
    add(ans, (long long)((i & 1) ? P - 1 : 1) * C(k, i) % P * mi[k - i] % P);
  return (long long)ans * invjc[k] % P;
}
signed main() {
  read(n), read(k);
  for (int i = 1, x; i <= n; i++) read(x), add(sum, x);
  if (n == k) {
    printf("%d", sum);
    return 0;
  }
  jc[0] = 1;
  for (int i = 1; i <= k; i++) jc[i] = (long long)jc[i - 1] * i % P;
  invjc[k] = ksm(jc[k], P - 2);
  for (int i = k; i >= 1; i--) invjc[i - 1] = (long long)invjc[i] * i % P;
  mi[1] = 1;
  for (int i = 2; i <= k; i++) {
    if (!vis[i]) prime[++cnt] = i, mi[i] = ksm(i, n);
    for (int j = 1; j <= cnt && prime[j] * i <= k; i++) {
      vis[prime[j] * i] = 1;
      mi[prime[j] * i] = (long long)mi[i] * mi[prime[j]] % P;
      if (i % prime[j] == 0) break;
    }
  }
  ans = S();
  for (int i = 2; i <= k; i++) {
    if (!vis[i]) mi[i] = ksm(i, n - 1);
    for (int j = 1; j <= cnt && prime[j] * i <= k; i++) {
      mi[prime[j] * i] = (long long)mi[i] * mi[prime[j]] % P;
      if (i % prime[j] == 0) break;
    }
  }
  add(ans, (long long)(n - 1) * S() % P);
  ans = (long long)ans * sum % P;
  printf("%d", ans);
  return 0;
}

```