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 double pi = 3.14159265359;
const int MAX = 1e6 + 2;
const long long int inf = 1e18;
const int N = 1010;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
const long long int mod = 1000000007;
inline long long int add(long long int a, long long int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline long long int mul(long long int a, long long int b) {
  return (((a + mod) % mod) * (b + mod) % mod) % mod;
}
inline long long int power(long long int a, long long int b) {
  long long int rt = 1;
  while (b > 0) {
    if (b & 1) rt = mul(rt, a);
    a = mul(a, a);
    b >>= 1;
  }
  return rt;
}
inline long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2) % p;
}
void stArt() {}
void eNd() {}
long long int i, j, k, l, m, r, n;
int main() {
  stArt();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int a, b;
  cin >> a >> b;
  long long int ans = a * (a + 1) / 2;
  ans %= mod;
  ans = mul(b, ans);
  ans = add(ans, a);
  k = b * (b - 1) / 2;
  k %= mod;
  ans = mul(ans, k);
  cout << ans;
  eNd();
  return 0;
}
