Celebrating the new year, many people post videos of falling dominoes; Here's a list of them: https://www.youtube.com/results?search_query=New+Years+Dominos User ainta, who lives in a 2D world, is going to post a video as well. There are n dominoes on a 2D Cartesian plane. i -th domino ( 1 ≤ i ≤ n ) can be represented as a line segment which is parallel to the y -axis and whose length is l i . The lower point of the domino is on the x -axis. Let's denote the x -coordinate of the i -th domino as p i . Dominoes are placed one after another, so p 1 < p 2 < ... < p n - 1 < p n holds. User ainta wants to take a video of falling dominoes. To make dominoes fall, he can push a single domino to the right. Then, the domino will fall down drawing a circle-shaped orbit until the line segment totally overlaps with the x-axis. Also, if the s -th domino touches the t -th domino while falling down, the t -th domino will also fall down towards the right, following the same procedure above. Domino s touches domino t if and only if the segment representing s and t intersects. See the picture above. If he pushes the leftmost domino to the right, it falls down, touching dominoes (A), (B) and (C). As a result, dominoes (A), (B), (C) will also fall towards the right. However, domino (D) won't be affected by pushing the leftmost domino, but eventually it will fall because it is touched by domino (C) for the first time. The picture above is an example of falling dominoes. Each red circle denotes a touch of two dominoes. User ainta has q plans of posting the video. j -th of them starts with pushing the x j -th domino, and lasts until the y j -th domino falls. But sometimes, it could be impossible to achieve such plan, so he has to lengthen some dominoes. It costs one dollar to increase the length of a single domino by 1 . User ainta wants to know, for each plan, the minimum cost needed to achieve it. Plans are processed independently, i. e. if domino's length is increased in some p