#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 51, MOD = 1e9 + 7;
int num, sum, cnt, k, res;
int fact[MAXN], finv[MAXN];
inline int read() {
  register int num = 0, neg = 1;
  register char ch = getchar();
  while (!isdigit(ch) && ch != '-') {
    ch = getchar();
  }
  if (ch == '-') {
    neg = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = (num << 3) + (num << 1) + (ch - '0');
    ch = getchar();
  }
  return num * neg;
}
inline int qpow(int base, int exponent) {
  int res = 1;
  while (exponent) {
    if (exponent & 1) {
      res = (long long int)res * base % MOD;
    }
    base = (long long int)base * base % MOD, exponent >>= 1;
  }
  return res;
}
inline void setup(int num) {
  fact[0] = finv[0] = 1;
  for (register int i = 1; i <= num; i++) {
    fact[i] = (long long int)fact[i - 1] * i % MOD;
  }
  finv[num] = qpow(fact[num], MOD - 2);
  for (register int i = num - 1; i; i--) {
    finv[i] = (long long int)finv[i + 1] * (i + 1) % MOD;
  }
}
inline int getSt(int n, int m) {
  long long int res = 0;
  for (register int i = 0; i <= n; i++) {
    res = res + (long long int)((n - i) & 1 ? MOD - 1 : 1) * qpow(i, m) % MOD *
                    finv[i] % MOD * finv[n - i] % MOD;
    res %= MOD;
  }
  return res;
}
int main() {
  cnt = read(), k = read(), setup(cnt);
  for (register int i = 0; i < cnt; i++) {
    sum = (sum + (num = read())) % MOD;
  }
  res = (getSt(k, cnt) + (long long int)(cnt - 1) * getSt(k, cnt - 1) % MOD) %
        MOD;
  printf("%d\n", (long long int)res * sum % MOD);
}

### Prompt

Please formulate 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 = 2e5 + 51, MOD = 1e9 + 7;
int num, sum, cnt, k, res;
int fact[MAXN], finv[MAXN];
inline int read() {
  register int num = 0, neg = 1;
  register char ch = getchar();
  while (!isdigit(ch) && ch != '-') {
    ch = getchar();
  }
  if (ch == '-') {
    neg = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = (num << 3) + (num << 1) + (ch - '0');
    ch = getchar();
  }
  return num * neg;
}
inline int qpow(int base, int exponent) {
  int res = 1;
  while (exponent) {
    if (exponent & 1) {
      res = (long long int)res * base % MOD;
    }
    base = (long long int)base * base % MOD, exponent >>= 1;
  }
  return res;
}
inline void setup(int num) {
  fact[0] = finv[0] = 1;
  for (register int i = 1; i <= num; i++) {
    fact[i] = (long long int)fact[i - 1] * i % MOD;
  }
  finv[num] = qpow(fact[num], MOD - 2);
  for (register int i = num - 1; i; i--) {
    finv[i] = (long long int)finv[i + 1] * (i + 1) % MOD;
  }
}
inline int getSt(int n, int m) {
  long long int res = 0;
  for (register int i = 0; i <= n; i++) {
    res = res + (long long int)((n - i) & 1 ? MOD - 1 : 1) * qpow(i, m) % MOD *
                    finv[i] % MOD * finv[n - i] % MOD;
    res %= MOD;
  }
  return res;
}
int main() {
  cnt = read(), k = read(), setup(cnt);
  for (register int i = 0; i < cnt; i++) {
    sum = (sum + (num = read())) % MOD;
  }
  res = (getSt(k, cnt) + (long long int)(cnt - 1) * getSt(k, cnt - 1) % MOD) %
        MOD;
  printf("%d\n", (long long int)res * sum % MOD);
}

```