#include <bits/stdc++.h>
using namespace std;
const int MOD = int(1e9) + 7;
inline int norm(int a) {
  while (a >= MOD) a -= MOD;
  while (a < 0) a += MOD;
  return a;
}
inline int mul(int a, int b) { return int((a * 1ll * b) % MOD); }
inline int binPow(int a, int k) {
  int ans = 1;
  while (k > 0) {
    if (k & 1) ans = mul(ans, a);
    a = mul(a, a);
    k >>= 1;
  }
  return ans;
}
const int N = 200 * 1000 + 555;
int f[N], inf[N];
void precalc() {
  f[0] = inf[0] = 1;
  for (int i = int(1); i < int(N); i++) {
    f[i] = mul(f[i - 1], i);
    inf[i] = mul(inf[i - 1], binPow(i, MOD - 2));
  }
}
int n, k, w[N];
inline bool read() {
  if (!(cin >> n >> k)) return false;
  for (int i = int(0); i < int(n); i++) cin >> w[i];
  return true;
}
inline int c(int n, int k) {
  if (k > n || n < 0) return 0;
  if (k == 0 || n == k) return 1;
  return mul(f[n], mul(inf[n - k], inf[k]));
}
inline int s(int n, int k) {
  if (n == 0) return k == 0;
  if (k == 0) return n == 0;
  int ans = 0, sg[2] = {1, MOD - 1};
  for (int cnt = int(0); cnt < int(k); cnt++)
    ans = norm(ans + mul(sg[cnt & 1], mul(c(k, cnt), binPow(k - cnt, n))));
  return mul(ans, inf[k]);
}
inline void solve() {
  int sum = 0;
  for (int i = int(0); i < int(n); i++) sum = norm(sum + w[i]);
  int s0 = s(n, k);
  int s1 = mul(n - 1, s(n - 1, k));
  cout << mul(sum, norm(s0 + s1)) << endl;
}
int main() {
  precalc();
  assert(read());
  solve();
  return 0;
}

### Prompt

Your challenge is to write 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 MOD = int(1e9) + 7;
inline int norm(int a) {
  while (a >= MOD) a -= MOD;
  while (a < 0) a += MOD;
  return a;
}
inline int mul(int a, int b) { return int((a * 1ll * b) % MOD); }
inline int binPow(int a, int k) {
  int ans = 1;
  while (k > 0) {
    if (k & 1) ans = mul(ans, a);
    a = mul(a, a);
    k >>= 1;
  }
  return ans;
}
const int N = 200 * 1000 + 555;
int f[N], inf[N];
void precalc() {
  f[0] = inf[0] = 1;
  for (int i = int(1); i < int(N); i++) {
    f[i] = mul(f[i - 1], i);
    inf[i] = mul(inf[i - 1], binPow(i, MOD - 2));
  }
}
int n, k, w[N];
inline bool read() {
  if (!(cin >> n >> k)) return false;
  for (int i = int(0); i < int(n); i++) cin >> w[i];
  return true;
}
inline int c(int n, int k) {
  if (k > n || n < 0) return 0;
  if (k == 0 || n == k) return 1;
  return mul(f[n], mul(inf[n - k], inf[k]));
}
inline int s(int n, int k) {
  if (n == 0) return k == 0;
  if (k == 0) return n == 0;
  int ans = 0, sg[2] = {1, MOD - 1};
  for (int cnt = int(0); cnt < int(k); cnt++)
    ans = norm(ans + mul(sg[cnt & 1], mul(c(k, cnt), binPow(k - cnt, n))));
  return mul(ans, inf[k]);
}
inline void solve() {
  int sum = 0;
  for (int i = int(0); i < int(n); i++) sum = norm(sum + w[i]);
  int s0 = s(n, k);
  int s1 = mul(n - 1, s(n - 1, k));
  cout << mul(sum, norm(s0 + s1)) << endl;
}
int main() {
  precalc();
  assert(read());
  solve();
  return 0;
}

```