#include <bits/stdc++.h>
const double pi = acos(-1);
const long long inf = 1e9 + 7;
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f * x;
}
int n, m;
long long ans, s, p[200005], inv[200005], invp[200005];
long long qpow(long long x, long long t) {
  long long s = 1;
  for (; t; t >>= 1, x = ((x) * (x)) % inf)
    if (t & 1) s = s * x % inf;
  return s;
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) ans += read(), ans %= inf;
  p[0] = p[1] = invp[1] = invp[0] = inv[1] = 1;
  for (int i = 2; i <= m; i++)
    inv[i] = inv[inf % i] * (inf - inf / i) % inf,
    invp[i] = invp[i - 1] * inv[i] % inf, p[i] = p[i - 1] * i % inf;
  for (int k = 0; k <= m; k++) {
    long long t = qpow(m - k, n - 1);
    if (k & 1)
      s += inf - invp[k] * invp[m - k] % inf * t % inf * (m - k) % inf,
          s %= inf;
    else
      s += invp[k] * invp[m - k] % inf * t % inf * (m - k) % inf, s %= inf;
    if (k & 1)
      s += inf - invp[k] * invp[m - k] % inf * t % inf * (n - 1) % inf,
          s %= inf;
    else
      s += invp[k] * invp[m - k] % inf * t % inf * (n - 1) % inf, s %= inf;
  }
  return 0 * printf("%lld\n", ans * s % inf);
}

### 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>
const double pi = acos(-1);
const long long inf = 1e9 + 7;
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f * x;
}
int n, m;
long long ans, s, p[200005], inv[200005], invp[200005];
long long qpow(long long x, long long t) {
  long long s = 1;
  for (; t; t >>= 1, x = ((x) * (x)) % inf)
    if (t & 1) s = s * x % inf;
  return s;
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) ans += read(), ans %= inf;
  p[0] = p[1] = invp[1] = invp[0] = inv[1] = 1;
  for (int i = 2; i <= m; i++)
    inv[i] = inv[inf % i] * (inf - inf / i) % inf,
    invp[i] = invp[i - 1] * inv[i] % inf, p[i] = p[i - 1] * i % inf;
  for (int k = 0; k <= m; k++) {
    long long t = qpow(m - k, n - 1);
    if (k & 1)
      s += inf - invp[k] * invp[m - k] % inf * t % inf * (m - k) % inf,
          s %= inf;
    else
      s += invp[k] * invp[m - k] % inf * t % inf * (m - k) % inf, s %= inf;
    if (k & 1)
      s += inf - invp[k] * invp[m - k] % inf * t % inf * (n - 1) % inf,
          s %= inf;
    else
      s += invp[k] * invp[m - k] % inf * t % inf * (n - 1) % inf, s %= inf;
  }
  return 0 * printf("%lld\n", ans * s % inf);
}

```