Dreamoon loves summing up something for no reason. One day he obtains two integers a and b occasionally. He wants to calculate the sum of all nice integers. Positive integer x is called nice if <image> and <image>, where k is some integer number in range [1, a].

By <image> we denote the quotient of integer division of x and y. By <image> we denote the remainder of integer division of x and y. You can read more about these operations here: http://goo.gl/AcsXhT.

The answer may be large, so please print its remainder modulo 1 000 000 007 (109 + 7). Can you compute it faster than Dreamoon?

Input

The single line of the input contains two integers a, b (1 ≤ a, b ≤ 107).

Output

Print a single integer representing the answer modulo 1 000 000 007 (109 + 7).

Examples

Input

1 1


Output

0


Input

2 2


Output

8

Note

For the first sample, there are no nice integers because <image> is always zero.

For the second sample, the set of nice integers is {3, 5}.
#include <bits/stdc++.h>
using namespace std;
const int OO = 1e9;
const long long INF = 1e18;
const int irand(int lo, int hi) {
  return ((double)rand() / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
const long long lrand(long long lo, long long hi) {
  return ((double)rand() / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
const long long MOD = (long long)1e9 + 7;
long long addmod(long long a, long long b) { return (a + b) % MOD; }
long long mulmod(long long a, long long b) { return (a * b) % MOD; }
long long exp(long long a, long long e) {
  if (e == 1)
    return a;
  else if (e == 0)
    return 1;
  else {
    long long ret = exp(a, e / 2);
    ret = mulmod(ret, ret);
    if (e & 1) ret = mulmod(ret, a);
    return ret;
  }
}
long long divmod(long long a, long long b) {
  return mulmod(a, exp(b, MOD - 2));
}
long long getsum(long long x) { return divmod(mulmod(x, x + 1), 2); }
int main() {
  ios_base::sync_with_stdio(0);
  long long a, b;
  cin >> a >> b;
  if (b == 1)
    cout << 0 << endl;
  else {
    long long ans = 0;
    ans = addmod(mulmod(getsum(a), mulmod(b, getsum(b - 1))),
                 mulmod(getsum(b - 1), a));
    cout << ans << endl;
  }
  return 0;
}
