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 <iostream>
#include <cstdlib>
#include <cstdio>
using namespace std;
typedef bool boolean;

int X, n, q;
int *rs;
boolean down = true;

typedef class Data {
	public:
		int L, R;
		int low;

		Data() {	}

		void pass(int t, boolean down) {
			if (down) {
				L -= t, R -= t;
				if (L < 0)
					low += -L, L = 0, low = min(low, X);
				if (R < 0)
					R = 0;
			} else {
				L += t, R += t;
				if (R > X)
					R = X;
				if (L > X)
					low = X, L = X;
			}
		}

		int getVal(int v) {
			if (v < low)
				return L;
			if (v > low + R - L)
				return R;
			return L + v - low;
		}
}Data;

Data d;
inline void solve() {
	scanf("%d", &X);
	scanf("%d", &n);
	rs = new int[(n + 1)];
	for (int i = 1; i <= n; i++)
		scanf("%d", rs + i);
	scanf("%d", &q);
	
	int cur = 0, pr = 1, tc, tv;
	d.L = 0, d.R = X, d.low = 0;
	
	while (q--) {
		scanf("%d%d", &tc, &tv);
		while (pr <= n && rs[pr] <= tc) {
			d.pass(rs[pr] - cur, down);
			down = !down, cur = rs[pr], pr++;
		}
		d.pass(tc - cur, down), cur = tc;
		printf("%d\n", d.getVal(tv));
	}
}

int main() {
	solve();
	return 0;
}