There are four natural numbers 1, a, b, c, satisfying a+b+c=2001, and 1<a<b<c, these four small natural numbers can be summed in pairs to get 6 different numbers. Arrange these 6 numbers from small to large, find the difference between two adjacent numbers, and get 5 numbers, these 5 numbers are all equal, then C=how many.
有四个自然数1、a、b、c，满足a+b+c=2001，且1＜a＜b＜c，这四个小自然数两两求和可得到6个不同的数．把这6个数从小到大排列，求相邻两数的差，又得到5个数，这5个数恰都相等，则C=多少．
<result>889</result>