给定一个整数数组A，N我们N在2D平面中绘制圆盘，使得第i个圆盘的中心为(0,i)，半径为A[i]。我们说，如果第k和第j光盘具有至少一个公共点，则第k光盘和第j光盘相交。

写一个函数

int number_of_disc_intersections(int[] A);

给定一个如上所述的A描述N光盘的数组，它返回相交光盘对的数量。例如，给定N=6和

A[0] = 1
A[1] = 5
A[2] = 2
A[3] = 1
A[4] = 4
A[5] = 0

有11对相交的光盘：

0th and 1st
0th and 2nd
0th and 4th
1st and 2nd
1st and 3rd
1st and 4th
1st and 5th
2nd and 3rd
2nd and 4th
3rd and 4th
4th and 5th

因此函数应返回11。如果相交对的数量超过10,000,000，则函数应返回-1。该功能可以假定N不超过10,000,000。