This is the hard 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 $1 \le i \le n - 1$, 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 $i \neq n$, 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

The first line contains two integers $n$ and $q$ ($2 \le n \le 5\cdot 10^5$, $1 \le q \le 5\cdot 10^5$) — the number of water towers and the number of updates.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of liters of water in water tower $i$.

The third line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($0 \le b_i \le 10^9$) — the power of the wizard in front of water tower $i$.

The fourth line contains $n - 1$ integers $c_1, c_2, \ldots, c_{n - 1}$ ($0 \le c_i \color{red}{\le} 10^{18}$) — 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 $z$ ($1 \le p \le n$, $0 \le x, y \le 10^9$, $0 \le z \color{red}{\le} 10^{18}$) — 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

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. Even though there are $2$ liters of water remaining, only $1$ liter of water can flow into tower 4.
• When $i = 4$, there are $4$ liters of water in tower 4. All $4$ liters of water are turned into wine.

Hence, $W(a,b,c)=1 + 4 + 2 + 4 = 11$ after the first update.

The second update modifies the arrays to $a = [3, 5, 3, 3]$, $b = [1, 1, 2, 8]$, and $c = [5, 1, 1]$.

• 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. Even though there are $6$ liters of water remaining, only $1$ liter of water can flow to tower 3.
• When $i = 3$, there are $4$ liters of water in tower 3 and $2$ liters of water is turned into wine. Even though there are $2$ liters of water remaining, only $1$ liter of water can flow into tower 4.
• When $i = 4$, there are $4$ liters of water in tower 4. All $4$ liters of water are turned into wine.

Hence, $W(a,b,c)=1 + 1 + 2 + 4 = 8$ after the second update.

The third update modifies the arrays to $a = [3, 5, 0, 3]$, $b = [1, 1, 0, 8]$, and $c = [5, 1, 0]$.

• 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. Even though there are $6$ liters of water remaining, only $1$ liter of water can flow to tower 3.
• When $i = 3$, there is $1$ liter of water in tower 3 and $0$ liters of water is turned into wine. Even though there is $1$ liter of water remaining, no water can flow to tower 4.
• When $i = 4$, there are $3$ liters of water in tower 4. All $3$ liters of water are turned into wine.

Hence, $W(a,b,c)=1 + 1 + 0 + 3 = 5$ after the third update.