#include <bits/stdc++.h>
using namespace std;
const int maxn = 200007;
const int mod = 1e9 + 7;
long long pown(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
int add(int x, int y) { return (x + y) % mod; }
int mul(int x, int y) { return 1LL * x * y % mod; }
int inv[maxn], fac[maxn];
int C(int a, int b) {
  if (a < b) return 0;
  return mul(fac[a], mul(inv[b], inv[a - b]));
}
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int a[maxn];
int main() {
  fac[0] = inv[0] = 1;
  int n = read(), K = read();
  int tota = 0;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    tota = add(tota, a[i]);
  }
  for (int i = 1; i <= K; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  inv[K] = pown(fac[K], mod - 2);
  for (int i = K - 1; i; i--) inv[i] = mul(inv[i + 1], i + 1);
  int ans = 0;
  if (n == 1) {
    if (K == 1)
      printf("%d", tota);
    else
      printf("0");
    return 0;
  }
  for (int i = 0; i <= K - 1; i++) {
    int x = pown(mod - 1, i);
    x = mul(x, mul(inv[i], inv[K - 1 - i]));
    int y = K - 1 - i;
    int r = (mul(n, pown(y + 1, n - 1)) -
             mul(mul(n - 1, y), pown(y + 1, n - 2)) + mod) %
            mod;
    r = mul(r, x);
    ans = add(r, ans);
  }
  ans = mul(ans, tota);
  cout << ans;
  return 0;
}

### Prompt

Generate 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;
const int maxn = 200007;
const int mod = 1e9 + 7;
long long pown(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
int add(int x, int y) { return (x + y) % mod; }
int mul(int x, int y) { return 1LL * x * y % mod; }
int inv[maxn], fac[maxn];
int C(int a, int b) {
  if (a < b) return 0;
  return mul(fac[a], mul(inv[b], inv[a - b]));
}
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int a[maxn];
int main() {
  fac[0] = inv[0] = 1;
  int n = read(), K = read();
  int tota = 0;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    tota = add(tota, a[i]);
  }
  for (int i = 1; i <= K; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  inv[K] = pown(fac[K], mod - 2);
  for (int i = K - 1; i; i--) inv[i] = mul(inv[i + 1], i + 1);
  int ans = 0;
  if (n == 1) {
    if (K == 1)
      printf("%d", tota);
    else
      printf("0");
    return 0;
  }
  for (int i = 0; i <= K - 1; i++) {
    int x = pown(mod - 1, i);
    x = mul(x, mul(inv[i], inv[K - 1 - i]));
    int y = K - 1 - i;
    int r = (mul(n, pown(y + 1, n - 1)) -
             mul(mul(n - 1, y), pown(y + 1, n - 2)) + mod) %
            mod;
    r = mul(r, x);
    ans = add(r, ans);
  }
  ans = mul(ans, tota);
  cout << ans;
  return 0;
}

```