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;

const int N = 100005;

struct Info{ int s, e, x, y; };

int n, A, t[N], q;
Info a[N];

int main(){
    scanf("%d%d", &A, &n);
    a[0] = {0, A, 0, 0};
    for(int i = 1; i <= n; i++){
        scanf("%d", t + i);
        int dt = t[i] - t[i - 1];
        auto &p = a[i - 1];
        int ns = p.s, ne = p.e;
        if(~i & 1){ ns += dt; ne += dt; }
        else{ ns -= dt; ne -= dt; }
        if(ns >= A) a[i] = {A, A, 0, A};
        else if(ne > A) a[i] = {ns, A, p.x, p.y + ne - A};
        else if(ne <= 0) a[i] = {0, 0, A, 0};
        else if(ns < 0) a[i] = {0, ne, p.x - ns, p.y};
        else a[i] = {ns, ne, p.x, p.y};
    }
    scanf("%d", &q);
    for(int x, y, z; q--; ){
        scanf("%d%d", &x, &y);
        z = int(upper_bound(t, t + n + 1, x) - t - 1);
        y = max(a[z].s, min(a[z].e, a[z].s + y - a[z].x));
        if(z & 1) y = min(A, y + x - t[z]);
        else y = max(0, y - x + t[z]);
        printf("%d\n", y);
    }
}