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 main(){
    int X, K;
    cin >> X >> K;
    vector<int> R(K+2);
    for(int i=1; i<=K; i++) cin >> R[i];
    R[K+1] = 1e9;
    int Q;
    cin >> Q;
    vector<int> T(Q), A(Q);
    for(int i=0; i<Q; i++) cin >> T[i] >> A[i];

    int q = 0;
    int mni = 0, mn = 0, mxi = X, mx = X;
    bool rev = false;
    for(int i=0; i<=K; i++){
        //cerr << mni << " " << mn << " " << mxi << " " << mx << endl;
        while(q < Q && T[q] <= R[i+1]){
            int a = (rev ? X-A[q] : A[q]), t = T[q] - R[i];
            int ans = max(min(mx, max(mn, mn+a-mni)) - t, 0);
            if(rev) ans = X-ans;
            cout << ans << endl;
            q++;
        }
        mn -= R[i+1] - R[i];
        mx -= R[i+1] - R[i];
        if(mx <= 0){
            mni = mxi = mn = mx = 0;
        }else if(mn <= 0){
            mni -= mn;
            mn = 0;
        }
        mni = X-mni;
        mxi = X-mxi;
        mn  = X-mn;
        mx  = X-mx;
        swap(mni, mxi);
        swap(mn, mx);
        rev = !rev;
    }
}