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 ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++i)
using namespace std;
template <typename tn> void read (tn & a) {
	tn x = 0, f = 1; char c = getchar();
	while (c < '0' || c > '9') { if (c == '-') f = 0; c = getchar(); }
	while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
	a = f ? x : -x;
}
const int MAXN = 1e5 + 10;
int x, k, q;
int r[MAXN];
int sign, kx;
void calc (ll &k) {
	if(k < 0) k = 0;
	if(k > x) k = x;
}


int main () {
//	freopen("in.txt","r",stdin);
	read(x);
	read(k);
	rep(i, 1, k) read(r[i]);
	read(q);
	ll L = 0, R = x, kx = 1, m = 0;
	sign = -1;
	rep(i, 1, q) {
		int a, t;
		read(t);
		read(a);
		while (kx <= k && r[kx] <= t) {
			ll vx = (r[kx] - r[kx - 1]) * sign;
			R += vx;
			L += vx;
			m += vx;
			calc(R);
			calc(m);
			++kx;
			sign = -sign;
		}
		ll an = a + L;
		ll hh = (t - r[kx - 1]) * sign;
//		cerr<<" ]  "<<kx<<" "<<i<<" "<<t<<" "<<r[kx - 1]<<"\n";
		if (an < m) an = m;
		if (an > R) an = R;
		an += hh;
		calc(an);
//		cerr<<"! "<<kx<<" "<<L<<" "<<R<<" "<<m<<" "<<hh<<"\n";
		printf("%lld\n", an);
	}
	
	return 0;
}