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>
using namespace std;
#define db(x) cout << #x << "=" << x << "  "
#define el cout << endl;
#define re(i, n) for(int i = 0; i < n; ++i)
const int MXN = 2e5 + 10;

int X, D, U, K, Q;
int R[MXN], T[MXN], A[MXN];

int clip(int x, int low = 0, int high = X) {
    return max(low, min(x, high));
}

int delta = 0, tim = 0;
void adv(int t) {
    delta += t, tim += abs(t);
    U = clip(U + t);
    D = clip(D + t);
}

int main() {
    cin >> X >> K;
    re(i, K) scanf("%d", R + i);
    cin >> Q;
    D = 0, U = X;
    int p = 0;
    re(i, Q) {
        scanf("%d %d", T + i, A + i);
        while (p < K && R[p] < T[i]) {
            adv((R[p] - tim) * ((p & 1) * 2 - 1));
            ++p;
        }
        adv((T[i] - tim) * ((p & 1) * 2 - 1));
        printf("%d\n", clip(A[i] + delta, D, U));
    }
    return 0;
}