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 <vector>
#include <algorithm>
#include <iostream>
using namespace std;

class Operation {
    int l, r, d;
public:
    Operation() : l(0), r(1<<30), d(0) {};
    Operation(int l, int r, int d) : l(l), r(r), d(d) {}
    int apply(int a) {
        return max(l, min(r, a + d));
    };
    Operation append(Operation op) {
        if (r + op.d <= op.l) return Operation(op.l, op.l, d + op.d);
        else if (l + op.d >= op.r) return Operation(op.r, op.r, d + op.d);
        else return Operation(max(l + op.d, op.l), min(r + op.d, op.r), d + op.d);
    }
};

Operation operator*(Operation lhs, Operation rhs) {
    return lhs.append(rhs);
}

int main() {
    int x, k; cin >> x >> k;
    int r[k]; for (int i = 0; i < k; i++) cin >> r[i];
    vector<Operation> ops;
    ops.push_back(Operation(0, x, 0));
    for (int i = 1; i <= k; i++) {
        int r1 = i == 1 ? 0 : r[i-2], r2 = r[i-1];
        ops.push_back(ops[i-1] * Operation(0, x, (r2 - r1) * (i % 2 ? -1 : 1)));
    }
    int q; cin >> q;
    while (q--) {
        int t, a; cin >> t >> a;
        int i = upper_bound(r, r+k, t) - r;
        int a1 = ops[i].apply(a);
        cout << Operation(0, x, (t - (i > 0 ? r[i-1] : 0)) * (i % 2 ? 1 : -1)).apply(a1) << endl;
    }
}
