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 <cstdio>
#include <algorithm>

#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)
#define RepG(i, x) for (int i = head[x]; i; i = edge[i].next)

using namespace std;

typedef long long LL;
const int N = 100010;

LL r[N], a[N], b[N], z[N];
int main()
{
	LL x, k;
	scanf("%lld%lld", &x, &k);
	a[0] = 0, b[0] = x;
	Rep(i, k) {
		scanf("%lld", &r[i]);
		LL tmp = r[i] - r[i - 1];
		a[i] = a[i - 1], b[i] = b[i - 1], z[i] = z[i - 1];
		if (i & 1) {
			if (z[i] >= tmp) z[i] -= tmp;
			else if (tmp <= b[i] - a[i] + z[i])
				a[i] += tmp - z[i], z[i] = 0; 
			else z[i] = 0, a[i] = 0, b[i] = 0;
		}
		else {
			if (z[i] + b[i] - a[i] + tmp <= x) z[i] += tmp;
			else if (z[i] + tmp <= x)
				z[i] += tmp, b[i] -= z[i] + b[i] - a[i] - x;
			else z[i] = x, a[i] = 0, b[i] = 0;
		}
		//printf("%lld %lld %lld\n", a[i], b[i], z[i]);
	}
	int q;
	scanf("%d", &q);
	Rep(i, q) {
		LL a0, t0;
		scanf("%lld%lld", &t0, &a0);
		LL tmp = (upper_bound(r + 1, r + 1 + k, t0) - r) - 1;
		LL ans;
		if (a0 <= a[tmp]) ans = z[tmp];
		else if (a0 <= b[tmp]) ans = z[tmp] + a0 - a[tmp];
		else ans = z[tmp] + b[tmp] - a[tmp];
		LL tx = t0 - r[tmp];
		if ((tmp + 1) & 1) ans = max(0ll, ans - tx);
		else ans = min(x, ans + tx);
		
		printf("%lld\n", ans);
	}
	
	return 0;
}
/*
50 3
9 35 67
1
92 28
38

*/