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 N 200000
typedef long long LL;
using namespace std;
int r[N], X, k, Q;

int calc(LL x) {
    if (x > X) x = X;
    if (x < 0) x = 0;
    return x;
}

int main() {
    cin >> X >> k;
    for(int i = 1; i <= k; i++) scanf("%d", &r[i]);
    cin >> Q;
    int p = 1, L = 0, R = X, ty = -1;
    LL now = 0;
    while (Q--) {
        int t, a;scanf("%d%d", &t, &a);
        while (p <= k && r[p] <= t) {
            int tmp = (r[p] - r[p - 1]) * ty;
            L = calc((LL)L + tmp), R = calc((LL)R + tmp), now = now + tmp;
            p++, ty *= -1;
        }
        int det = t - r[p - 1], ret = now + a;
        if (ret > R) ret = R;
        if (ret < L) ret = L;
        ret = calc((LL)ret + det * ty);
        printf("%d\n", ret);
    }
}
