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>

const int maxn = 1e5 + 3;
int X, K, rev[maxn], Q, ti[maxn], ai[maxn], l, r, u, v;

void trans(int dt) {
	if(u + dt < 0) {
		l -= u + dt;
		u = 0;
	} else if(u + dt > X) {
		u = X;
		l = X;
	} else {
		u += dt;
	}
	if(v + dt < 0) {
		v = 0;
		r = 0;
	} else if(v + dt > X) {
		r -= v + dt - X;
		v = X;
	} else {
		v += dt;
	}
}

int calc(int x) {
	if(x <= l)
		return u;
	if(x >= r)
		return v;
	return u + x - l;
}

int main() {
	scanf("%d%d", &X, &K);
	l = u = 0;
	r = v = X;
	rev[0] = 0;
	for(int i = 1; i <= K; ++i)
		scanf("%d", rev + i);
	scanf("%d", &Q);
	for(int i = 0; i < Q; ++i)
		scanf("%d%d", ti + i, ai + i);
	ti[Q] = 0x7fffffff;
	for(int i = 1, j = 0, t = rev[0]; i <= K || j < Q;) {
		if(i <= K && rev[i] <= ti[j]) {
			trans((i & 1) ? t - rev[i] : rev[i] - t);
			t = rev[i++];
		} else {
			trans((i & 1) ? t - ti[j] : ti[j] - t);
			printf("%d\n", calc(ai[j]));
			t = ti[j++];
		}
	}
	return 0;
}
