#include <bits/stdc++.h>
inline int mod(int x) { return x < 1000000007 ? x : x - 1000000007; }
inline int calc(int x, int y) {
  int z = 1;
  for (; y; x = (long long)x * x % 1000000007, y >>= 1)
    if (y & 1) z = (long long)x * z % 1000000007;
  return z;
}
int a[200002], b[200002], i, j, k, m, n, x, y, z;
bool o[200002];
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &j), x = mod(j + x);
  if (n == 1) return 0 & printf("%d\n", x);
  for (a[1] = 1, i = 2; i <= m; i++) {
    if (!o[i]) a[b[k++] = i] = calc(i, n - 2);
    for (j = 0; i * b[j] <= m; j++) {
      o[i * b[j]] = 1, a[i * b[j]] = (long long)a[i] * a[b[j]] % 1000000007;
      if (i % b[j] == 0) break;
    }
  }
  for (*b = b[1] = 1, i = 2; i < m; i++)
    b[i] = (long long)(1000000007 - 1000000007 / i) * b[1000000007 % i] %
           1000000007;
  for (i = 2; i < m; i++) b[i] = (long long)b[i - 1] * b[i] % 1000000007;
  for (i = 0; i < m; i++)
    y = (y + (n - 1ll + m - i) * (i & 1 ? 1000000007 - a[m - i] : a[m - i]) %
                 1000000007 * b[i] % 1000000007 * b[m - 1 - i]) %
        1000000007;
  return 0 & printf("%lld\n", (long long)x * y % 1000000007);
}

### 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>
inline int mod(int x) { return x < 1000000007 ? x : x - 1000000007; }
inline int calc(int x, int y) {
  int z = 1;
  for (; y; x = (long long)x * x % 1000000007, y >>= 1)
    if (y & 1) z = (long long)x * z % 1000000007;
  return z;
}
int a[200002], b[200002], i, j, k, m, n, x, y, z;
bool o[200002];
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &j), x = mod(j + x);
  if (n == 1) return 0 & printf("%d\n", x);
  for (a[1] = 1, i = 2; i <= m; i++) {
    if (!o[i]) a[b[k++] = i] = calc(i, n - 2);
    for (j = 0; i * b[j] <= m; j++) {
      o[i * b[j]] = 1, a[i * b[j]] = (long long)a[i] * a[b[j]] % 1000000007;
      if (i % b[j] == 0) break;
    }
  }
  for (*b = b[1] = 1, i = 2; i < m; i++)
    b[i] = (long long)(1000000007 - 1000000007 / i) * b[1000000007 % i] %
           1000000007;
  for (i = 2; i < m; i++) b[i] = (long long)b[i - 1] * b[i] % 1000000007;
  for (i = 0; i < m; i++)
    y = (y + (n - 1ll + m - i) * (i & 1 ? 1000000007 - a[m - i] : a[m - i]) %
                 1000000007 * b[i] % 1000000007 * b[m - 1 - i]) %
        1000000007;
  return 0 & printf("%lld\n", (long long)x * y % 1000000007);
}

```