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 N = 100054;

int V, n, q, t;
int k1, k2, k3;
int rev[N];
bool Up = false;

inline void up(int &x, const int y) {x < y ? x = y : 0;}
inline void down(int &x, const int y) {x > y ? x = y : 0;}
inline int min(const int x, const int y) {return x < y ? x : y;}
inline int max(const int x, const int y) {return x < y ? y : x;}
inline int f(int x) {return min(max(x, k1), k2) + k3;} // clip
inline void handle(int I) {Up ? (k3 += I, down(k1, V - k3), down(k2, V - k3)) : (k3 -= I, up(k1, -k3), up(k2, -k3)), t += I;}

int main() {
	int i, tg, x;
	scanf("%d%d", &V, &n);
	for (i = 0; i < n; ++i) scanf("%d", rev + i);
	rev[n] = INT_MAX;
	i = 0, k2 = V;
	for (scanf("%d", &q); q; --q) {
		scanf("%d%d", &tg, &x);
		for (; rev[i] <= tg; ++i) handle(rev[i] - t), Up = !Up;
		handle(tg - t), printf("%d\n", f(x));
	}
	return 0;
}