#include <bits/stdc++.h>
using namespace std;
const int N = 200001;
const long long mod = 1000000007ll;
int n, k;
long long fact[N], inv[N], inv_fact[N];
long long w[N], s = 0;
long long C(int a, int b) {
  if (a < b)
    return 0;
  else
    return (fact[a] * ((inv_fact[b] * inv_fact[a - b]) % mod)) % mod;
}
long long qpow(long long a, int m) {
  long long ret = 1;
  while (m) {
    if (m & 1) ret = (ret * a) % mod;
    a = (a * a) % mod, m >>= 1;
  }
  return ret;
}
long long S(int a, int b) {
  long long total = 0;
  if (a < b) return 0;
  for (int i = 0; i <= b; i++) {
    long long delta = 0;
    delta = C(b, i);
    if ((b - i) & 1) delta = (mod - delta) % mod;
    total = (total + delta * qpow(1ll * i, a)) % mod;
  }
  return (total * inv_fact[b]) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%lld", &w[i]), s = (s + w[i]) % mod;
  inv[1] = 1;
  for (int i = 2; i <= n; i++)
    inv[i] = ((mod / i) * (mod - inv[mod % i])) % mod;
  fact[0] = 1, inv_fact[0] = 1;
  for (int i = 1; i <= n; i++)
    fact[i] = (fact[i - 1] * 1ll * i) % mod,
    inv_fact[i] = (inv_fact[i - 1] * inv[i]) % mod;
  long long num = (S(n, k) + 1ll * (n - 1) * S(n - 1, k)) % mod;
  printf("%lld\n", (num * s) % mod);
  return 0;
}

### Prompt

Please formulate 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 N = 200001;
const long long mod = 1000000007ll;
int n, k;
long long fact[N], inv[N], inv_fact[N];
long long w[N], s = 0;
long long C(int a, int b) {
  if (a < b)
    return 0;
  else
    return (fact[a] * ((inv_fact[b] * inv_fact[a - b]) % mod)) % mod;
}
long long qpow(long long a, int m) {
  long long ret = 1;
  while (m) {
    if (m & 1) ret = (ret * a) % mod;
    a = (a * a) % mod, m >>= 1;
  }
  return ret;
}
long long S(int a, int b) {
  long long total = 0;
  if (a < b) return 0;
  for (int i = 0; i <= b; i++) {
    long long delta = 0;
    delta = C(b, i);
    if ((b - i) & 1) delta = (mod - delta) % mod;
    total = (total + delta * qpow(1ll * i, a)) % mod;
  }
  return (total * inv_fact[b]) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%lld", &w[i]), s = (s + w[i]) % mod;
  inv[1] = 1;
  for (int i = 2; i <= n; i++)
    inv[i] = ((mod / i) * (mod - inv[mod % i])) % mod;
  fact[0] = 1, inv_fact[0] = 1;
  for (int i = 1; i <= n; i++)
    fact[i] = (fact[i - 1] * 1ll * i) % mod,
    inv_fact[i] = (inv_fact[i - 1] * inv[i]) % mod;
  long long num = (S(n, k) + 1ll * (n - 1) * S(n - 1, k)) % mod;
  printf("%lld\n", (num * s) % mod);
  return 0;
}

```