#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long n, k;
long long sum;
long long f[200000];
template <typename T>
T inv(T a, T b) {
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
long long pow(long long a, long long k) {
  long long ret = 1LL, s = a;
  while (k) {
    if (k & 1) ret = (ret * s) % mod;
    s = (s * s) % mod;
    k >>= 1;
  }
  return ret;
}
long long fact(long long n) {
  if (f[n] == 0LL) f[n] = (n * fact(n - 1)) % mod;
  return f[n];
}
long long stirling(long long n, long long k) {
  long long ret = 0LL;
  for (long long j = 1; j <= k; j++) {
    long long cur =
        (pow(j, n - 1) * inv((fact(j - 1) * fact(k - j)) % mod, mod)) % mod;
    ret += ((k ^ j) & 1) ? -cur : cur;
  }
  ret %= mod;
  return ret < 0 ? ret + mod : ret;
}
int main() {
  f[0] = 1LL;
  scanf("%lld %lld", &n, &k);
  for (int i = 0; i < n; i++) {
    long long tmp;
    scanf("%lld", &tmp);
    sum += tmp;
  }
  sum %= mod;
  if (n == 1LL) {
    printf("%lld\n", sum);
    return 0;
  }
  long long ans = (stirling(n, k) + ((n - 1) * stirling(n - 1, k)) % mod) % mod;
  ans = (ans * sum) % mod;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 long long mod = 1000000007;
long long n, k;
long long sum;
long long f[200000];
template <typename T>
T inv(T a, T b) {
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
long long pow(long long a, long long k) {
  long long ret = 1LL, s = a;
  while (k) {
    if (k & 1) ret = (ret * s) % mod;
    s = (s * s) % mod;
    k >>= 1;
  }
  return ret;
}
long long fact(long long n) {
  if (f[n] == 0LL) f[n] = (n * fact(n - 1)) % mod;
  return f[n];
}
long long stirling(long long n, long long k) {
  long long ret = 0LL;
  for (long long j = 1; j <= k; j++) {
    long long cur =
        (pow(j, n - 1) * inv((fact(j - 1) * fact(k - j)) % mod, mod)) % mod;
    ret += ((k ^ j) & 1) ? -cur : cur;
  }
  ret %= mod;
  return ret < 0 ? ret + mod : ret;
}
int main() {
  f[0] = 1LL;
  scanf("%lld %lld", &n, &k);
  for (int i = 0; i < n; i++) {
    long long tmp;
    scanf("%lld", &tmp);
    sum += tmp;
  }
  sum %= mod;
  if (n == 1LL) {
    printf("%lld\n", sum);
    return 0;
  }
  long long ans = (stirling(n, k) + ((n - 1) * stirling(n - 1, k)) % mod) % mod;
  ans = (ans * sum) % mod;
  printf("%lld\n", ans);
  return 0;
}

```