CF734C.Anton and Making Potions

Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next level he has to prepare n potions.

Anton has a special kettle, that can prepare one potions in x seconds. Also, he knows spells of two types that can faster the process of preparing potions.

1. Spells of this type speed up the preparation time of one potion. There are m spells of this type, the i-th of them costs b__i manapoints and changes the preparation time of each potion to a__i instead of x.
2. Spells of this type immediately prepare some number of potions. There are k such spells, the i-th of them costs d__i manapoints and instantly create c__i potions.

Anton can use no more than one spell of the first type and no more than one spell of the second type, and the total number of manapoints spent should not exceed s. Consider that all spells are used instantly and right before Anton starts to prepare potions.

Anton wants to get to the next level as fast as possible, so he is interested in the minimum number of time he needs to spent in order to prepare at least n potions.

安东正在玩一款非常有趣的电脑游戏，但目前他卡在了其中一关。为了进入下一关，他必须制作 n 瓶药水。

安东有一个特殊的烧瓶，可以在 x 秒内制作一瓶药水。此外，他还掌握了两类法术，可加快药水的制作过程。

1. 此类法术可缩短单瓶药水的制作时间。共有 m 个此类法术，其中第 i 个法术消耗 b__i 点法力值，并将每瓶药水的制作时间由 x 秒变为 a__i 秒。
2. 此类法术可立即制作若干瓶药水。共有 k 个此类法术，其中第 i 个法术消耗 d__i 点法力值，并立即生成 c__i 瓶药水。

安东最多只能使用一个第一类法术和一个第二类法术，且所消耗的总法力值不得超过 s。假设所有法术均瞬间生效，且在安东开始制作药水之前立即使用。

安东希望尽快通关，因此他关心的是：为制作至少 n 瓶药水所需的最短时间。

The first line of the input contains three integers n, m, k (1 ≤ n ≤ 2·109, 1 ≤ m, k ≤ 2·105) — the number of potions, Anton has to make, the number of spells of the first type and the number of spells of the second type.

The second line of the input contains two integers x and s (2 ≤ x ≤ 2·109, 1 ≤ s ≤ 2·109) — the initial number of seconds required to prepare one potion and the number of manapoints Anton can use.

The third line contains m integers a__i (1 ≤ a__i < x) — the number of seconds it will take to prepare one potion if the i-th spell of the first type is used.

The fourth line contains m integers b__i (1 ≤ b__i ≤ 2·109) — the number of manapoints to use the i-th spell of the first type.

There are k integers c__i (1 ≤ c__i ≤ n) in the fifth line — the number of potions that will be immediately created if the i-th spell of the second type is used. It's guaranteed that c__i are not decreasing, i.e. c__i ≤ c__j if i < j.

The sixth line contains k integers d__i (1 ≤ d__i ≤ 2·109) — the number of manapoints required to use the i-th spell of the second type. It's guaranteed that d__i are not decreasing, i.e. d__i ≤ d__j if i < j.

输入的第一行包含三个整数 $n$、$m$、$k$（$1 ≤ n ≤ 2·10^9$，$1 ≤ m,\,k ≤ 2·10^5$）——分别表示 Anton 需要制作的药水数量、第一类法术的数量以及第二类法术的数量。

输入的第二行包含两个整数 $x$ 和 $s$（$2 ≤ x ≤ 2·10^9$，$1 ≤ s ≤ 2·10^9$）——分别表示初始制作一瓶药水所需的时间（秒数）以及 Anton 可用的法力值（manapoints）。

输入的第三行包含 $m$ 个整数 $a_i$（$1 ≤ a_i < x$）——若使用第 $i$ 个第一类法术，则制作一瓶药水所需的时间（秒数）。

输入的第四行包含 $m$ 个整数 $b_i$（$1 ≤ b_i ≤ 2·10^9$）——使用第 $i$ 个第一类法术所需的法力值。

输入的第五行包含 $k$ 个整数 $c_i$（$1 ≤ c_i ≤ n$）——若使用第 $i$ 个第二类法术，则立即生成的药水数量。保证 $c_i$ 非递减，即当 $i < j$ 时，有 $c_i ≤ c_j$。

输入的第六行包含 $k$ 个整数 $d_i$（$1 ≤ d_i ≤ 2·10^9$）——使用第 $i$ 个第二类法术所需的法力值。保证 $d_i$ 非递减，即当 $i < j$ 时，有 $d_i ≤ d_j$。

Print one integer — the minimum time one has to spent in order to prepare n potions.

输出一个整数——制备 n 瓶药水所需的最少时间。

In the first sample, the optimum answer is to use the second spell of the first type that costs 10 manapoints. Thus, the preparation time of each potion changes to 4 seconds. Also, Anton should use the second spell of the second type to instantly prepare 15 potions spending 80 manapoints. The total number of manapoints used is 10 + 80 = 90, and the preparation time is 4·5 = 20 seconds (15 potions were prepared instantly, and the remaining 5 will take 4 seconds each).

In the second sample, Anton can't use any of the spells, so he just prepares 20 potions, spending 10 seconds on each of them and the answer is 20·10 = 200.

在第一个样例中，最优方案是使用第一类法术中的第二个，其消耗 10 点法力值。因此，每瓶药水的准备时间变为 4 秒。此外，Anton 还应使用第二类法术中的第二个，以瞬间制备 15 瓶药水，消耗 80 点法力值。总共消耗的法力值为 $10 + 80 = 90$，总准备时间为 $4 \cdot 5 = 20$ 秒（其中 15 瓶药水瞬间完成，剩余 5 瓶每瓶需耗时 4 秒）。

在第二个样例中，Anton 无法使用任何法术，因此他只能直接制备 20 瓶药水，每瓶耗时 10 秒，答案为 $20 \cdot 10 = 200$。