#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/STACK:16777216")
using namespace std;
using matrix = vector<vector<long long>>;
void FAST_IO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long mod = 1e9 + 7;
const int MAXN = 2e5 + 100;
long long fact[MAXN], rfact[MAXN];
long long binpow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    n >>= 1;
  }
  return res;
}
long long c(int n, int k) {
  long long res = fact[n];
  res = (res * rfact[k]) % mod;
  res = (res * rfact[n - k]) % mod;
  return res;
}
long long s(int n, int k) {
  long long res = 0;
  for (int i = 0; i <= k; ++i) {
    long long tmp = (c(k, i) * binpow(i, n)) % mod;
    if ((k + i) & 1) {
      tmp = mod - tmp;
    }
    res = (res + tmp) % mod;
  }
  res = (res * rfact[k]) % mod;
  return res;
}
signed main() {
  FAST_IO();
  fact[0] = rfact[0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    fact[i] = (fact[i - 1] * i) % mod;
    rfact[i] = binpow(fact[i], mod - 2);
  }
  int n, k;
  cin >> n >> k;
  long long coeff = (s(n, k) + (n - 1) * s(n - 1, k)) % mod;
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    ans = (ans + x * coeff) % mod;
  }
  cout << ans;
}

### 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>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/STACK:16777216")
using namespace std;
using matrix = vector<vector<long long>>;
void FAST_IO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long mod = 1e9 + 7;
const int MAXN = 2e5 + 100;
long long fact[MAXN], rfact[MAXN];
long long binpow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    n >>= 1;
  }
  return res;
}
long long c(int n, int k) {
  long long res = fact[n];
  res = (res * rfact[k]) % mod;
  res = (res * rfact[n - k]) % mod;
  return res;
}
long long s(int n, int k) {
  long long res = 0;
  for (int i = 0; i <= k; ++i) {
    long long tmp = (c(k, i) * binpow(i, n)) % mod;
    if ((k + i) & 1) {
      tmp = mod - tmp;
    }
    res = (res + tmp) % mod;
  }
  res = (res * rfact[k]) % mod;
  return res;
}
signed main() {
  FAST_IO();
  fact[0] = rfact[0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    fact[i] = (fact[i - 1] * i) % mod;
    rfact[i] = binpow(fact[i], mod - 2);
  }
  int n, k;
  cin >> n >> k;
  long long coeff = (s(n, k) + (n - 1) * s(n - 1, k)) % mod;
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    ans = (ans + x * coeff) % mod;
  }
  cout << ans;
}

```