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>

#define mod 1000000007
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define nyan "(=^･ω･^=)"
#define mkp make_pair
#define lP pair<ll, ll>
#define iP pair<int,int>

using namespace std;
typedef long long ll;

int X, K, A[100002], Q, t, a, p, q, l, r, L, R;

int main() {
	cin >> X >> K;
	for (int i = 1; i != K + 1; ++i)
		cin >> A[i];
	A[K + 1] = intmax;
	cin >> Q;
	r = X;
	R = X;
	while (Q--) {
		cin >> t >> a;
		for (; p != K + 1; ++p) {
			if (A[p + 1] >= t) {
				int res;
				if (p & 1) {
					if (a < l) {
						res = L + t - A[p];
					}
					else if (a <= r)
						res = a + q + t - A[p];
					else
						res = R + t - A[p];
					res = min(res, X);
				}
				else {
					if (a < l) {
						res = L - t + A[p];
					}
					else if (a <= r)
						res = a + q - t + A[p];
					else
						res = R - t + A[p];
					res = max(res, 0);
				}
				cout << res << endl;
				break;
			}
			if (p & 1) {
				L += A[p + 1] - A[p];
				q += A[p + 1] - A[p];
				R += A[p + 1] - A[p];
				L = min(L, X);
				R = min(R, X);
				r = min(r, X - q);

			}
			else {
				L -= A[p + 1] - A[p];
				q -= A[p + 1] - A[p];
				R -= A[p + 1] - A[p];
				L = max(L, 0);
				R = max(R, 0);
				l = max(l, -q);
			}
		}
	}
	return 0;
}