#include <bits/stdc++.h>
using namespace std;
ifstream in("test.in");
ofstream out("test.out");
const int DIM = 2e5 + 9;
const int MOD = 1e9 + 7;
int fct[DIM], inv[DIM];
int logPow(int x, int n) {
  if (!n) {
    return 1;
  }
  int y = logPow(x, n / 2);
  if (n | 1) {
    y = 1LL * y * y % MOD;
  }
  if (n & 1) {
    y = 1LL * y * x % MOD;
  }
  return y;
}
inline int f(int x) {
  if (x < 0) {
    return x + MOD;
  };
  if (x >= MOD) {
    return x - MOD;
  }
  return x;
}
inline int getCombinations(int n, int k) {
  return 1LL * fct[n] * inv[k] % MOD * inv[n - k] % MOD;
}
inline int getStirling(int n, int k) {
  int ans = 0;
  for (int i = 0, v = (k & 1) ? -1 : 1; i <= k; ++i, v = -v) {
    ans = f(ans + (1LL * getCombinations(k, i) * logPow(i, n) % MOD) * v);
  }
  return (int)(1LL * ans * inv[k] % MOD);
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  fct[0] = inv[0] = 1;
  for (int i = 1; i <= k; ++i) {
    fct[i] = 1LL * fct[i - 1] * i % MOD;
  }
  inv[k] = logPow(fct[k], MOD - 2);
  for (int i = k - 1; i >= 1; --i) {
    inv[i] = 1LL * inv[i + 1] * (i + 1) % MOD;
  }
  int sum = 0;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    sum = f(sum + x);
  }
  int ans = (1LL * (n - 1) * getStirling(n - 1, k) + getStirling(n, k)) % MOD;
  cout << 1LL * ans * sum % MOD << endl;
  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;
ifstream in("test.in");
ofstream out("test.out");
const int DIM = 2e5 + 9;
const int MOD = 1e9 + 7;
int fct[DIM], inv[DIM];
int logPow(int x, int n) {
  if (!n) {
    return 1;
  }
  int y = logPow(x, n / 2);
  if (n | 1) {
    y = 1LL * y * y % MOD;
  }
  if (n & 1) {
    y = 1LL * y * x % MOD;
  }
  return y;
}
inline int f(int x) {
  if (x < 0) {
    return x + MOD;
  };
  if (x >= MOD) {
    return x - MOD;
  }
  return x;
}
inline int getCombinations(int n, int k) {
  return 1LL * fct[n] * inv[k] % MOD * inv[n - k] % MOD;
}
inline int getStirling(int n, int k) {
  int ans = 0;
  for (int i = 0, v = (k & 1) ? -1 : 1; i <= k; ++i, v = -v) {
    ans = f(ans + (1LL * getCombinations(k, i) * logPow(i, n) % MOD) * v);
  }
  return (int)(1LL * ans * inv[k] % MOD);
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  fct[0] = inv[0] = 1;
  for (int i = 1; i <= k; ++i) {
    fct[i] = 1LL * fct[i - 1] * i % MOD;
  }
  inv[k] = logPow(fct[k], MOD - 2);
  for (int i = k - 1; i >= 1; --i) {
    inv[i] = 1LL * inv[i + 1] * (i + 1) % MOD;
  }
  int sum = 0;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    sum = f(sum + x);
  }
  int ans = (1LL * (n - 1) * getStirling(n - 1, k) + getStirling(n, k)) % MOD;
  cout << 1LL * ans * sum % MOD << endl;
  return 0;
}

```