#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
inline int add(int x, int y) {
  x += y;
  return x >= mod ? x - mod : x;
}
inline int sub(int x, int y) {
  x -= y;
  return x < 0 ? x + mod : x;
}
inline int mul(long long x, int y) {
  x *= y;
  return x >= mod ? x % mod : x;
}
inline int mpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = mul(ret, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ret;
}
int fac[202020], ifac[202020];
inline int C(int a, int b) { return mul(fac[a], mul(ifac[b], ifac[a - b])); }
inline int S(int n, int k) {
  if (n < k) return 0;
  int ret = 0;
  for (int i = 1; i <= k; i++) {
    int tmp = C(k, i);
    tmp = mul(tmp, mpow(i, n));
    if ((k - i) & 1)
      ret = sub(ret, tmp);
    else
      ret = add(ret, tmp);
  }
  return mul(ret, ifac[k]);
}
int n, k;
int main() {
  fac[0] = ifac[0] = 1;
  for (int i = 1; i < 202020; i++) {
    fac[i] = mul(fac[i - 1], i);
    ifac[i] = mpow(fac[i], mod - 2);
  }
  scanf("%d%d", &n, &k);
  int bns = add(S(n, k), mul(n - 1, S(n - 1, k)));
  int sum = 0;
  for (int i = 0, x; i < n; i++) {
    scanf("%d", &x);
    sum = add(sum, x);
  }
  printf("%d\n", mul(sum, bns));
}

### Prompt

Your task is to create 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 mod = 1000000007;
inline int add(int x, int y) {
  x += y;
  return x >= mod ? x - mod : x;
}
inline int sub(int x, int y) {
  x -= y;
  return x < 0 ? x + mod : x;
}
inline int mul(long long x, int y) {
  x *= y;
  return x >= mod ? x % mod : x;
}
inline int mpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = mul(ret, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ret;
}
int fac[202020], ifac[202020];
inline int C(int a, int b) { return mul(fac[a], mul(ifac[b], ifac[a - b])); }
inline int S(int n, int k) {
  if (n < k) return 0;
  int ret = 0;
  for (int i = 1; i <= k; i++) {
    int tmp = C(k, i);
    tmp = mul(tmp, mpow(i, n));
    if ((k - i) & 1)
      ret = sub(ret, tmp);
    else
      ret = add(ret, tmp);
  }
  return mul(ret, ifac[k]);
}
int n, k;
int main() {
  fac[0] = ifac[0] = 1;
  for (int i = 1; i < 202020; i++) {
    fac[i] = mul(fac[i - 1], i);
    ifac[i] = mpow(fac[i], mod - 2);
  }
  scanf("%d%d", &n, &k);
  int bns = add(S(n, k), mul(n - 1, S(n - 1, k)));
  int sum = 0;
  for (int i = 0, x; i < n; i++) {
    scanf("%d", &x);
    sum = add(sum, x);
  }
  printf("%d\n", mul(sum, bns));
}

```