#include <bits/stdc++.h>
using namespace std;
inline long long LL(int x) { return static_cast<long long>(x); }
namespace MOD_BASE {
const int Dollar1 = 1000000007;
inline int up(int x) { return x >= Dollar1 ? x - Dollar1 : x; }
inline int down(int x) { return x < 0 ? x + Dollar1 : x; }
inline int& add(int& x, const int y) {
  x += y;
  if (x >= Dollar1) x -= Dollar1;
  return x;
}
inline int& dec(int& x, const int y) {
  x -= y;
  if (x < 0) x += Dollar1;
  return x;
}
inline int mul(int x, int y) { return LL(x) * y % Dollar1; }
inline int fastpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = LL(res) * a % Dollar1;
    a = LL(a) * a % Dollar1;
    b >>= 1;
  }
  return res;
}
inline int INV(int x) { return fastpow(x, Dollar1 - 2); }
int fac[1000010], inv[1000010], invfac[1000010];
inline int C(int a, int b) {
  if (a < 0 || b < 0 || b > a) return 0;
  return LL(fac[a]) * invfac[b] % Dollar1 * invfac[a - b] % Dollar1;
}
inline int lucas(int a, int b) {
  if (a < Dollar1 && b < Dollar1) return C(a, b);
  return LL(lucas(a % Dollar1, b % Dollar1)) * lucas(a / Dollar1, b / Dollar1) %
         Dollar1;
}
int RUN_INIT() {
  invfac[0] = invfac[1] = inv[0] = inv[1] = fac[0] = fac[1] = 1;
  for (int i = 2; i != 1000010; ++i) {
    fac[i] = LL(fac[i - 1]) * i % Dollar1;
    inv[i] = LL(inv[Dollar1 % i]) * (Dollar1 - Dollar1 / i) % Dollar1;
  }
  for (int i = 2; i != 1000010; ++i) invfac[i] = mul(invfac[i - 1], inv[i]);
  return 0;
}
int FIRSTRUN = RUN_INIT();
}  // namespace MOD_BASE
using namespace MOD_BASE;
int S(int n, int m) {
  int res = 0;
  for (int i = 0, t = 1; i <= m; ++i, t = -t)
    add(res, down(t * mul(fastpow(m - i, n), C(m, i))));
  return mul(res, invfac[m]);
}
int n, k, w[200010], cnt[200010];
int main() {
  scanf("%d%d", &n, &k);
  int cnt = 0;
  for (int i = 1; i <= n; ++i) scanf("%d", &w[i]), add(cnt, w[i]);
  printf("%d\n", mul(cnt, up(S(n, k) + mul(n - 1, S(n - 1, k)))));
  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;
inline long long LL(int x) { return static_cast<long long>(x); }
namespace MOD_BASE {
const int Dollar1 = 1000000007;
inline int up(int x) { return x >= Dollar1 ? x - Dollar1 : x; }
inline int down(int x) { return x < 0 ? x + Dollar1 : x; }
inline int& add(int& x, const int y) {
  x += y;
  if (x >= Dollar1) x -= Dollar1;
  return x;
}
inline int& dec(int& x, const int y) {
  x -= y;
  if (x < 0) x += Dollar1;
  return x;
}
inline int mul(int x, int y) { return LL(x) * y % Dollar1; }
inline int fastpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = LL(res) * a % Dollar1;
    a = LL(a) * a % Dollar1;
    b >>= 1;
  }
  return res;
}
inline int INV(int x) { return fastpow(x, Dollar1 - 2); }
int fac[1000010], inv[1000010], invfac[1000010];
inline int C(int a, int b) {
  if (a < 0 || b < 0 || b > a) return 0;
  return LL(fac[a]) * invfac[b] % Dollar1 * invfac[a - b] % Dollar1;
}
inline int lucas(int a, int b) {
  if (a < Dollar1 && b < Dollar1) return C(a, b);
  return LL(lucas(a % Dollar1, b % Dollar1)) * lucas(a / Dollar1, b / Dollar1) %
         Dollar1;
}
int RUN_INIT() {
  invfac[0] = invfac[1] = inv[0] = inv[1] = fac[0] = fac[1] = 1;
  for (int i = 2; i != 1000010; ++i) {
    fac[i] = LL(fac[i - 1]) * i % Dollar1;
    inv[i] = LL(inv[Dollar1 % i]) * (Dollar1 - Dollar1 / i) % Dollar1;
  }
  for (int i = 2; i != 1000010; ++i) invfac[i] = mul(invfac[i - 1], inv[i]);
  return 0;
}
int FIRSTRUN = RUN_INIT();
}  // namespace MOD_BASE
using namespace MOD_BASE;
int S(int n, int m) {
  int res = 0;
  for (int i = 0, t = 1; i <= m; ++i, t = -t)
    add(res, down(t * mul(fastpow(m - i, n), C(m, i))));
  return mul(res, invfac[m]);
}
int n, k, w[200010], cnt[200010];
int main() {
  scanf("%d%d", &n, &k);
  int cnt = 0;
  for (int i = 1; i <= n; ++i) scanf("%d", &w[i]), add(cnt, w[i]);
  printf("%d\n", mul(cnt, up(S(n, k) + mul(n - 1, S(n - 1, k)))));
  return 0;
}

```