CF677B Vanya and Food Processor

瓦尼亚在一个垂直的食品处理器中粉碎土豆。 你可以把它想象成一个圆柱体，从上面塞入，从下面粉碎后吐出。 每个土豆可以视为条状。 处理器中的土豆高度不超过$h$(否则会满出来)，处理器每秒粉碎$k$厘米的土豆。如果处理器里剩不到$k$厘米土豆，则粉碎所有剩余的土豆。 瓦尼亚有$n$条土豆，第$i$块的长度等于$a_i$。他把它们按顺序从$1$号到$n$号放进食品处理器，从$1$号开始，到$n$号结束。 每秒会发生如下事件： 1.如果还有至少一条土豆没放进去，瓦尼亚将它们逐一放入处理器，直到没有足够的空间放置下一片，即塞到塞不进为止。 2.处理器粉碎了$k$厘米或剩下全部的土豆。

输入的第一行包含整数$n$、$h$和$k$ （$1≤n≤100 000,1≤k≤h≤109$），含义如上。 第二行包含$n$个整数$a_i$（1≤a_i≤h），即土豆长度。

最短需要粉碎所有土豆的时间。

Consider the first sample. 1. First Vanya puts the piece of potato of height $ 5 $ into processor. At the end of the second there is only amount of height $ 2 $ remaining inside. 2. Now Vanya puts the piece of potato of height $ 4 $ . At the end of the second there is amount of height $ 3 $ remaining. 3. Vanya puts the piece of height $ 3 $ inside and again there are only $ 3 $ centimeters remaining at the end of this second. 4. Vanya finally puts the pieces of height $ 2 $ and $ 1 $ inside. At the end of the second the height of potato in the processor is equal to $ 3 $ . 5. During this second processor finally smashes all the remaining potato and the process finishes. In the second sample, Vanya puts the piece of height $ 5 $ inside and waits for $ 2 $ seconds while it is completely smashed. Then he repeats the same process for $ 4 $ other pieces. The total time is equal to $ 2·5=10 $ seconds. In the third sample, Vanya simply puts all the potato inside the processor and waits $ 2 $ seconds.