#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000 + 1000;
const int MOD = 1000000000 + 7;
int qpow(int x, int y) {
  int ret = 1;
  for (; y; x = (long long)x * x % MOD, y >>= 1)
    if (y & 1) ret = (long long)ret * x % MOD;
  return ret;
}
int n, k;
int a[MAXN], sum;
int fac[MAXN], ifac[MAXN];
int ans = 0;
void init() {
  fac[0] = 1;
  for (int i = 1; i < MAXN; i++) fac[i] = (long long)fac[i - 1] * i % MOD;
  ifac[MAXN - 1] = qpow(fac[MAXN - 1], MOD - 2);
  for (int i = MAXN - 2; i >= 0; i--)
    ifac[i] = (long long)ifac[i + 1] * (i + 1) % MOD;
}
int get_c(int n, int m) {
  if (n < 0 || m < 0 || n < m) return 0;
  return (long long)fac[n] * ifac[m] % MOD * ifac[n - m] % MOD;
}
int get_f(int x) {
  if (n == 0) return 0;
  if (n == 1) return 1;
  return (qpow(1 + x, n - 1) + (long long)(n - 1) * qpow(1 + x, n - 2)) % MOD;
}
int main() {
  ios::sync_with_stdio(false);
  init();
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sum = 0;
  for (int i = 1; i <= n; i++) sum = (sum + a[i]) % MOD;
  for (int i = 0; i <= k - 1; i++) {
    long long tmp = 1;
    tmp = (long long)tmp * ((k - 1 - i) & 1 ? MOD - 1 : 1) % MOD;
    tmp = (long long)tmp * get_c(k - 1, i) % MOD;
    tmp = (long long)tmp * get_f(i) % MOD;
    ans = (ans + tmp) % MOD;
  }
  ans = (long long)ans * ifac[k - 1] % MOD;
  ans = (long long)ans * sum % MOD;
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 = 200000 + 1000;
const int MOD = 1000000000 + 7;
int qpow(int x, int y) {
  int ret = 1;
  for (; y; x = (long long)x * x % MOD, y >>= 1)
    if (y & 1) ret = (long long)ret * x % MOD;
  return ret;
}
int n, k;
int a[MAXN], sum;
int fac[MAXN], ifac[MAXN];
int ans = 0;
void init() {
  fac[0] = 1;
  for (int i = 1; i < MAXN; i++) fac[i] = (long long)fac[i - 1] * i % MOD;
  ifac[MAXN - 1] = qpow(fac[MAXN - 1], MOD - 2);
  for (int i = MAXN - 2; i >= 0; i--)
    ifac[i] = (long long)ifac[i + 1] * (i + 1) % MOD;
}
int get_c(int n, int m) {
  if (n < 0 || m < 0 || n < m) return 0;
  return (long long)fac[n] * ifac[m] % MOD * ifac[n - m] % MOD;
}
int get_f(int x) {
  if (n == 0) return 0;
  if (n == 1) return 1;
  return (qpow(1 + x, n - 1) + (long long)(n - 1) * qpow(1 + x, n - 2)) % MOD;
}
int main() {
  ios::sync_with_stdio(false);
  init();
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sum = 0;
  for (int i = 1; i <= n; i++) sum = (sum + a[i]) % MOD;
  for (int i = 0; i <= k - 1; i++) {
    long long tmp = 1;
    tmp = (long long)tmp * ((k - 1 - i) & 1 ? MOD - 1 : 1) % MOD;
    tmp = (long long)tmp * get_c(k - 1, i) % MOD;
    tmp = (long long)tmp * get_f(i) % MOD;
    ans = (ans + tmp) % MOD;
  }
  ans = (long long)ans * ifac[k - 1] % MOD;
  ans = (long long)ans * sum % MOD;
  cout << ans << endl;
  return 0;
}

```