We have a sandglass consisting of two bulbs, bulb A and bulb B. These bulbs contain some amount of sand. When we put the sandglass, either bulb A or B lies on top of the other and becomes the upper bulb. The other bulb becomes the lower bulb.

The sand drops from the upper bulb to the lower bulb at a rate of 1 gram per second. When the upper bulb no longer contains any sand, nothing happens.

Initially at time 0, bulb A is the upper bulb and contains a grams of sand; bulb B contains X-a grams of sand (for a total of X grams).

We will turn over the sandglass at time r_1,r_2,..,r_K. Assume that this is an instantaneous action and takes no time. Here, time t refer to the time t seconds after time 0.

You are given Q queries. Each query is in the form of (t_i,a_i). For each query, assume that a=a_i and find the amount of sand that would be contained in bulb A at time t_i.

Constraints

* 1≤X≤10^9
* 1≤K≤10^5
* 1≤r_1<r_2< .. <r_K≤10^9
* 1≤Q≤10^5
* 0≤t_1<t_2< .. <t_Q≤10^9
* 0≤a_i≤X (1≤i≤Q)
* All input values are integers.

Input

The input is given from Standard Input in the following format:


X
K
r_1 r_2 .. r_K
Q
t_1 a_1
t_2 a_2
:
t_Q a_Q


Output

For each query, print the answer in its own line.

Examples

Input

180
3
60 120 180
3
30 90
61 1
180 180


Output

60
1
120


Input

100
1
100000
4
0 100
90 100
100 100
101 100


Output

100
10
0
0


Input

100
5
48 141 231 314 425
7
0 19
50 98
143 30
231 55
342 0
365 100
600 10


Output

19
52
91
10
58
42
100
#include <bits/stdc++.h>
typedef long long i64;
using std::cout;
using std::endl;
using std::cin;

int main(){
	int X, k, Q; cin >> X >> k; std::vector<int> r(k + 2, (1 << 30));
	for(int i = 0; i < k; i++) cin >> r[i + 1]; r[0] = 0;
	cin >> Q;
	
	int tap = 0, lis = X; // 潰れヴァリュー
	int low = 0, high = X; // 潰れ判定
	int cnt = 0, sum = 0;
	
	for(int i = 0; i < Q; i++) {
		int t, a; cin >> t >> a;
		
		while(r[cnt + 1] <= t) {
			int d = r[cnt + 1] - r[cnt];
			
			if(cnt % 2) {
				int uku = std::max(0, lis + d - X);
				high -= uku;
		
				tap = std::max(0, std::min(X, tap + d));
				lis = std::max(0, std::min(X, lis + d));
			} else {
				int luz = std::max(0, -(tap - d));
				low += luz;
				
				tap = std::max(0, std::min(X, tap - d));
				lis = std::max(0, std::min(X, lis - d));
			}
			
			cnt += 1;
		}
		
		int d = t - r[cnt];
		if(cnt % 2) {
			if(low <= a and a <= high) cout << std::max(0, std::min(X, tap + a - low + d)) << endl;
		 	else if(a < low) cout << std::max(0, std::min(X, tap + d)) << endl;
			else cout << std::max(0, std::min(X, lis + d)) << endl;
		} else {
			if(low <= a and a <= high) cout << std::max(0, std::min(X, tap + a - low - d)) << endl;
			else if(a < low) cout << std::max(0, std::min(X, tap - d)) << endl;
			else cout << std::max(0, std::min(X, lis - d)) << endl;	
		}
	}
	return 0;
}
