#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e6 + 100;
const int inf = 0x3f3f3f3f;
const int iinf = 1 << 30;
const long long linf = 2e18;
const long long mod = 1e9 + 7;
const double eps = 1e-7;
long long power(long long a, long long b) {
  long long ret = 1;
  for (; b; b >>= 1, (a *= a) %= mod)
    if (b & 1) (ret *= a) %= mod;
  return ret;
}
long long n, m;
long long a[maxn], fac[maxn], inv[maxn];
long long s1, s2, ans;
void init() {
  fac[0] = 1;
  for (register int i = (1); i <= (n); ++i) fac[i] = fac[i - 1] * i % mod;
  inv[n] = power(fac[n], mod - 2);
  for (register int i = (n); i >= (1); --i) inv[i - 1] = inv[i] * i % mod;
}
signed main() {
  scanf("%lld %lld", &n, &m);
  init();
  for (register int i = (1); i <= (n); ++i) scanf("%lld", &a[i]);
  for (register int i = (0); i <= (m); ++i) {
    (s1 += power(mod - 1, i) * power(m - i, n) % mod * inv[i] % mod *
           inv[m - i] % mod) %= mod;
    (s2 += power(mod - 1, i) * power(m - i, n - 1) % mod * inv[i] % mod *
           inv[m - i] % mod) %= mod;
  }
  for (register int i = (1); i <= (n); ++i)
    (ans += a[i] * (s1 + (n - 1) * s2 % mod) % mod) %= mod;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate 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 maxn = 2e6 + 100;
const int inf = 0x3f3f3f3f;
const int iinf = 1 << 30;
const long long linf = 2e18;
const long long mod = 1e9 + 7;
const double eps = 1e-7;
long long power(long long a, long long b) {
  long long ret = 1;
  for (; b; b >>= 1, (a *= a) %= mod)
    if (b & 1) (ret *= a) %= mod;
  return ret;
}
long long n, m;
long long a[maxn], fac[maxn], inv[maxn];
long long s1, s2, ans;
void init() {
  fac[0] = 1;
  for (register int i = (1); i <= (n); ++i) fac[i] = fac[i - 1] * i % mod;
  inv[n] = power(fac[n], mod - 2);
  for (register int i = (n); i >= (1); --i) inv[i - 1] = inv[i] * i % mod;
}
signed main() {
  scanf("%lld %lld", &n, &m);
  init();
  for (register int i = (1); i <= (n); ++i) scanf("%lld", &a[i]);
  for (register int i = (0); i <= (m); ++i) {
    (s1 += power(mod - 1, i) * power(m - i, n) % mod * inv[i] % mod *
           inv[m - i] % mod) %= mod;
    (s2 += power(mod - 1, i) * power(m - i, n - 1) % mod * inv[i] % mod *
           inv[m - i] % mod) %= mod;
  }
  for (register int i = (1); i <= (n); ++i)
    (ans += a[i] * (s1 + (n - 1) * s2 % mod) % mod) %= mod;
  printf("%lld\n", ans);
  return 0;
}

```