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;
struct edge {
  int t, f, w;
  bool operator<(const edge& e) const { return w > e.w; }
  edge(int from, int to, int weight) {
    t = to;
    f = from;
    w = weight;
  }
};
const double EPS = (1e-5);
struct line {
  double a, b, c;
};
struct point {
  double x, y;
  point() { x = y = 0.0; }
  point(double _x, double _y) : x(_x), y(_y) {}
  bool operator==(point other) const {
    return (fabs(x - other.x) < EPS && (fabs(y - other.y) < EPS));
  }
};
double dist(point p1, point p2) { return hypot(p1.x - p2.x, p1.y - p2.y); }
void pointsToLine(point p1, point p2, line& l) {
  if (fabs(p1.x - p2.x) < EPS) {
    l.a = 1.0;
    l.b = 0.0;
    l.c = -p1.x;
  } else {
    l.a = -(double)(p1.y - p2.y) / (p1.x - p2.x);
    l.b = 1.0;
    l.c = -(double)(l.a * p1.x) - p1.y;
  }
}
bool areParallel(line l1, line l2) {
  return (fabs(l1.a - l2.a) < EPS) && (fabs(l1.b - l2.b) < EPS);
}
bool areSame(line l1, line l2) {
  return areParallel(l1, l2) && (fabs(l1.c - l2.c) < EPS);
}
bool areIntersect(line l1, line l2, point& p) {
  if (areParallel(l1, l2)) return false;
  p.x = (l2.b * l1.c - l1.b * l2.c) / (l2.a * l1.b - l1.a * l2.b);
  if (fabs(l1.b) > EPS)
    p.y = -(l1.a * p.x + l1.c);
  else
    p.y = -(l2.a * p.x + l2.c);
  return true;
}
long long binpowmod(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int findXOR(int n) {
  switch (n % 4) {
    case 0:
      return n;
    case 1:
      return 1;
    case 2:
      return n + 1;
    case 3:
      return 0;
  }
}
int rangeXOR(int l, int r) { return (findXOR(l - 1) ^ findXOR(r)); }
int getbit(int mask, int bit) { return (mask & (1 << bit)); }
void setbit(int& mask, int bit, int val) {
  if (val)
    mask |= (1 << bit);
  else
    mask &= ~(1 << bit);
}
const int N = 1e6 + 10;
const int M = 26;
const long long INF = N;
long long sum(long long n) {
  long long a = n;
  long long b = n + 1;
  if (n & 1)
    b /= 2;
  else
    a /= 2;
  return (a * b) % 1000000007;
}
void solve() {
  long long a, b;
  cin >> a >> b;
  long long sumb = sum(b - 1);
  long long suma = sum(a);
  long long k = (b * suma) % 1000000007;
  k = (k + a) % 1000000007;
  long long ans = (sumb * k) % 1000000007;
  cout << ans << '\n';
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tc = 1;
  while (tc--) {
    solve();
  }
}
