Problem: You are given two sequences a and b, both of length 2N. The i-th elements in a and b are a_i and b_i, respectively. Using these sequences, Snuke is doing the job of calculating the beauty of pairs of balanced sequences of parentheses (defined below) of length 2N. The beauty of a pair (s,t) is calculated as follows:

* Let X=0.
* For each i between 1 and 2N (inclusive), increment X by a_i if s_i = t_i, and increment X by b_i otherwise.
* The beauty of (s,t) is the final value of X.



You will be given Q queries. Process them in order. In the i-th query, update the value of a_{p_i} to x_i, and the value of b_{p_i} to y_i. Then, find the maximum possible beauty of a pair of balanced sequences of parentheses.

In this problem, only the sequences below are defined to be balanced sequences of parentheses.

* An empty string
* The concatenation of `(`, s, `)` in this order, where s is a balanced sequence of parentheses
* The concatenation of s, t in this order, where s and t are balanced sequences of parentheses

Constraints

* 1 \leq N,Q \leq 10^{5}
* -10^{9} \leq a_i,b_i,x_i,y_i \leq 10^{9}
* 1 \leq p_i \leq 2N
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N Q
a_1 a_2 ... a_{2N}
b_1 b_2 ... b_{2N}
p_1 x_1 y_1
:
p_Q x_Q y_Q


Output

Print Q lines. The i-th line should contain the response to the i-th query.

Examples

Input

2 2
1 1 7 3
4 2 3 3
2 4 6
3 2 5


Output

15
15


Input

7 7
34 -20 -27 42 44 29 9 11 20 44 27 19 -31 -29
46 -50 -11 20 28 46 12 13 33 -22 -48 -27 35 -17
7 27 34
12 -2 22
4 -50 -12
3 -32 15
8 -7 23
3 -30 11
4 -2 23


Output

311
312
260
286
296
292
327

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
typedef long long ll;

using namespace std;

int N, Q;
ll A[200010], B[200010]; //[0, 2 * N) !!
multiset<ll> S;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> N >> Q;
	rep(i, 0, 2 * N) cin >> A[i];
	rep(i, 0, 2 * N) cin >> B[i];

	ll ans = 0; bool odd = false;
	ans += A[0] + A[2 * N - 1];
	rep(i, 1, 2 * N - 1) {
		ans += max(A[i], B[i]);
		odd ^= (A[i] > B[i]);
		S.insert(abs(A[i] - B[i]));
	}
	while(Q--) {
		ll a, x, y;
		cin >> a >> x >> y; a--;
		if(a == 0 || a == 2 * N - 1) {
			ans += x - A[a];
			A[a] = x;
		}
		else {
			odd ^= (A[a] > B[a]);
			odd ^= (x > y);
			ans += max(x, y) - max(A[a], B[a]);
			S.erase(S.lower_bound(abs(A[a] - B[a])));
			S.insert(abs(x - y));
			A[a] = x; B[a] = y;
		}
		cout << (!odd ? ans : ans - (*S.begin())) << "\n";
	}
}