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;
typedef long long ll;
const int K = 1e5+9;
ll r[K];
inline ll read() {
    ll s = 1, a = 0; char ch = getchar();
    while (ch < '0' || ch > '9') {if (ch == '-') s = -1; ch = getchar();}
    while (ch >= '0' && ch <= '9') {a = a * 10 + ch - '0'; ch = getchar();}
    return s * a;
}
inline void turn(ll &v, ll x) {
    if (v < 0) v = 0;
    if (v > x) v = x;
}
int main() {
    ll x = read(), k = read();
    for (int i = 1; i <= k; i++) r[i] = read();
    ll q = read();
    ll low = 0, mid = 0, upp = x;
    int i = 1, sign = -1;
    while (q--) {
        ll t = read(), a = read();
        while (i <= k && r[i] <= t) {
            ll dta = (r[i] - r[i - 1]) * sign;
            low += dta, upp += dta, mid += dta;
            turn(low, x), turn(upp, x);
            i++, sign *= -1;
        }
        ll res = a + mid;
        ll dta = (t - r[i - 1]) * sign;
        if (res < low) res = low;
        if (res > upp) res = upp;
        res += dta, turn(res, x);
        printf("%lld\n", res);
    }
    return 0;
}