Zxr960115 is owner of a large farm. He feeds m cute cats and employs p feeders. There's a straight road across the farm and n hills along the road, numbered from 1 to n from left to right. The distance between hill i and (i - 1) is di meters. The feeders live in hill 1.

ZXR是个农场主,他养了M只猫,雇了P只饲养员。农场中有一条笔直的路,路边有N座山,从1到N编号。第i座山与第i-1座山之间的距离是Di。饲养员都住在1号山。

One day, the cats went out to play. Cat i went on a trip to hill hi, finished its trip at time ti, and then waited at hill hi for a feeder. The feeders must take all the cats. Each feeder goes straightly from hill 1 to n without waiting at a hill and takes all the waiting cats at each hill away. Feeders walk at a speed of 1 meter per unit time and are strong enough to take as many cats as they want.

有一天,猫出去玩。第i只猫去Hi号山玩,玩到时刻Ti停止,然后在原地等待饲养员来接。饲养员们必须回收所有的猫。每个饲养员沿着路从1号山走到N号山,把各座山上已经在等待的猫全部接走。饲养员在路上行走需要时间,速度为1米/单位时间。饲养员在每座山上接猫的时间可以忽略,可以携带的猫的数量为无穷大。

For example, suppose we have two hills (d2 = 1) and one cat that finished its trip at time 3 at hill 2 (h1 = 2). Then if the feeder leaves hill 1 at time 2 or at time 3, he can take this cat, but if he leaves hill 1 at time 1 he can't take it. If the feeder leaves hill 1 at time 2, the cat waits him for 0 time units, if the feeder leaves hill 1 at time 3, the cat waits him for 1 time units.

例如有两座相距为1的山,一只猫在2号山玩,玩到时刻3开始等待。如果饲养员从1号山在时刻2或3出发,那么他可以接到猫,猫的等待时间为0或1。而如果他于时刻1出发,那么他将于时刻2经过2号山,不能接到当时仍在玩的猫。

Your task is to schedule the time (may be negative) leaving from hill 1 for each feeder so that the sum of the waiting time of all cats is minimized.

你的任务是规划每个饲养员从1号山出发的时间,使得所有猫等待时间的总和尽量小。饲养员出发的时间可以为负。

Input
The first line of the input contains three integers n, m, p (2 ≤ n ≤ 10^5, 1 ≤ m ≤ 10^5, 1 ≤ p ≤ 100).
The second line contains n - 1 positive integers d2, d3, ..., dn (1 ≤ di < 10^4).
Each of the next m lines contains two integers hi and ti (1 ≤ hi ≤ n, 0 ≤ ti ≤ 10^9).

Output
Output an integer, the minimum sum of waiting time of all cats.
Please, do not write the %I64d specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %lld specifier.

Sample test(s)
Input

4 6 2
1 3 5
1 0
2 1
4 9
1 10
2 10
3 12

Output
3