write a go solution for Description: This is the easy version of the problem. The only difference between the two versions is the constraint on c_i and z. You can make hacks only if both versions of the problem are solved. There are three arrays a, b and c. a and b have length n and c has length n-1. Let W(a,b,c) denote the liters of wine created from the following process. Create n water towers. The i-th water tower initially has a_i liters of water and has a wizard with power b_i in front of it. Furthermore, for each 1<=i<=n-1, there is a valve connecting water tower i to i+1 with capacity c_i. For each i from 1 to n in this order, the following happens: 1. The wizard in front of water tower i removes at most b_i liters of water from the tower and turns the removed water into wine. 2. If ineqn, at most c_i liters of the remaining water left in water tower i flows through the valve into water tower i+1. There are q updates. In each update, you will be given integers p, x, y and z and you will update a_p:=x, b_p:=y and c_p:=z. After each update, find the value of W(a,b,c). Note that previous updates to arrays a, b and c persist throughout future updates. Input Format: The first line contains two integers n and q (2<=n<=5*10^5, 1<=q<=5*10^5) — the number of water towers and the number of updates. The second line contains n integers a_1,a_2,ldots,a_n (0<=a_i<=10^9) — the number of liters of water in water tower i. The third line contains n integers b_1,b_2,ldots,b_n (0<=b_i<=10^9) — the power of the wizard in front of water tower i. The fourth line contains n-1 integers c_1,c_2,ldots,c_n-1 (c_icolorred=10^18) — the capacity of the pipe connecting water tower i to i+1. Each of the next q lines contains four integers p, x, y and z (1<=p<=n, 0<=x,y<=10^9, zcolorred=10^18) — the updates done to arrays a, b and c. Note that c_n does not exist, so the value of z does not matter when p=n. Output Format: Print q lines, each line containing a single integer representing W(a,b,c) after each update. Note: The first update does not make any modifications to the arrays. - When i=1, there are 3 liters of water in tower 1 and 1 liter of water is turned into wine. The remaining 2 liters of water flow into tower 2. - When i=2, there are 5 liters of water in tower 2 and 4 liters of water is turned into wine. The remaining 1 liter of water flows into tower 3. - When i=3, there are 4 liters of water in tower 3 and 2 liters of water is turned into wine. The remaining 2 liters of water flows into tower 4. - When i=4, there are 5 liters of water in tower 4. All 5 liters of water are turned into wine. Hence, W(a,b,c)=1+4+2+5=12 after the first update. The second update modifies the arrays to a=[3,5,3,3], b=[1,1,2,8], and c=[10^18,10^18,10^18]. - When i=1, there are 3 liters of water in tower 1 and 1 liter of water is turned into wine. The remaining 2 liters of water flow into tower 2. - When i=2, there are 7 liters of water in tower 2 and 1 liter of water is turned into wine. The remaining 6 liters of water flow into tower 3. - When i=3, there are 9 liters of water in tower 3 and 2 liters of water is turned into wine. The remaining 7 liters of water flow into tower 4. - When i=4, there are 10 liters of water in tower 4. Only 8 liters of water is turned into wine. Hence, W(a,b,c)=1+1+2+8=12 after the second update. The third update modifies the arrays to a=[3,5,0,3], b=[1,1,0,8], and c=[10^18,10^18,10^18]. - When i=1, there are 3 liters of water in tower 1 and 1 liter of water is turned into wine. The remaining 2 liters of water flow into tower 2. - When i=2, there are 7 liters of water in tower 2 and 1 liter of water is turned into wine. The remaining 6 liters of water flow into tower 3. - When i=3, there are 6 liters of water in tower 3 and 0 liters of water is turned into wine. The remaining 6 liters of water flow into tower 4. - When i=4, there are 9 liters of water in tower 4. Only 8 liters of water is turned into wine. Hence, W(a,b,c)=1+1+0+8=10 after the third update.. Output only the code with no comments, explanation, or additional text.