#include <bits/stdc++.h>
using namespace std;
const int N = 524288;
const int mod = 1e9 + 7;
int qpow(int x, int y = mod - 2) {
  int res = 1;
  for (; y; x = (long long)x * x % mod, y >>= 1)
    if (y & 1) res = (long long)res * x % mod;
  return res;
}
int fac[N], ifac[N];
int binom(int x, int y) {
  return (long long)fac[x] * ifac[y] % mod * ifac[x - y] % mod;
}
void math_init(int x) {
  fac[0] = 1;
  for (int i = 1, iE = x; i <= iE; i++)
    fac[i] = (long long)fac[i - 1] * i % mod;
  ifac[x] = qpow(fac[x]);
  for (int i = x, iE = 1; i >= iE; i--)
    ifac[i - 1] = (long long)ifac[i] * i % mod;
}
int fpow(int x) { return x % 2 ? mod - 1 : 1; }
int n, m, all, w[N], ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m, math_init(n);
  for (int i = 1, iE = n; i <= iE; i++) cin >> w[i], (all += w[i]) %= mod;
  if (m == 1) return cout << (long long)all * n % mod << endl, 0;
  for (int i = 0, iE = m - 1; i <= iE; i++) {
    (ans += (long long)fpow(m - 1 - i) * binom(m - 1, i) % mod *
            (qpow(i + 1, n - 1) % mod +
             (long long)(n - 1) * qpow(i + 1, n - 2) % mod) %
            mod) %= mod;
  }
  ans = (long long)ans * all % mod * ifac[m - 1] % 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 N = 524288;
const int mod = 1e9 + 7;
int qpow(int x, int y = mod - 2) {
  int res = 1;
  for (; y; x = (long long)x * x % mod, y >>= 1)
    if (y & 1) res = (long long)res * x % mod;
  return res;
}
int fac[N], ifac[N];
int binom(int x, int y) {
  return (long long)fac[x] * ifac[y] % mod * ifac[x - y] % mod;
}
void math_init(int x) {
  fac[0] = 1;
  for (int i = 1, iE = x; i <= iE; i++)
    fac[i] = (long long)fac[i - 1] * i % mod;
  ifac[x] = qpow(fac[x]);
  for (int i = x, iE = 1; i >= iE; i--)
    ifac[i - 1] = (long long)ifac[i] * i % mod;
}
int fpow(int x) { return x % 2 ? mod - 1 : 1; }
int n, m, all, w[N], ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m, math_init(n);
  for (int i = 1, iE = n; i <= iE; i++) cin >> w[i], (all += w[i]) %= mod;
  if (m == 1) return cout << (long long)all * n % mod << endl, 0;
  for (int i = 0, iE = m - 1; i <= iE; i++) {
    (ans += (long long)fpow(m - 1 - i) * binom(m - 1, i) % mod *
            (qpow(i + 1, n - 1) % mod +
             (long long)(n - 1) * qpow(i + 1, n - 2) % mod) %
            mod) %= mod;
  }
  ans = (long long)ans * all % mod * ifac[m - 1] % mod;
  cout << ans << endl;
  return 0;
}

```