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;

int r[100001];

int main(){
    #ifdef DEBUG
    freopen("input.txt", "r", stdin);
    #endif
    ios_base::sync_with_stdio(0), cin.tie(0);
    int x, k;
    cin >> x >> k;
    for (int i = 1; i <= k; i++){
        cin >> r[i];
    }
    int q;
    cin >> q;
    int j = 0;
    int d = 0;
    int mn = 0;
    int mx = x;
    int valmn = 0;
    int valmx = x;
    for (int i = 0; i < q; i++){
        int t1, a1;
        cin >> t1 >> a1;
        while (j + 1 <= k && t1 >= r[j + 1]){
            int ch = r[j + 1] - r[j];
            if (j & 1){
                mx = min(mx, mx + (x - valmx - ch));
                valmn = min(valmn + ch, x);
                valmx = min(valmx + ch, x);
            } else {
                mn = max(mn, mn + ch - valmn);
                valmn = max(valmn - ch, 0);
                valmx = max(valmx - ch, 0);
            }
            j++;
        }
        int ans;
        if (a1 <= mn){
            ans = valmn;
        } else if (a1 >= mx){
            ans = valmx;
        } else {
            ans = valmn + (a1 - mn);
        }
        if (j & 1) {
            ans = min(ans + (t1 - r[j]), x);
        } else {
            ans = max(ans - (t1 - r[j]), 0);
        }
        cout << ans << "\n";
    }
    return 0;
}

