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;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl





using ll = long long;

const ll mod = 1e9+7;
const int maxn = 1e6 + 5;



ll x, k, q;
ll r[maxn];
ll t[maxn], a[maxn];

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);

    cin>>x;
    cin>>k;
    for (int i=0; i<k; i++) {
	cin>>r[i];
    }

    cin>>q;
    for (int i=0; i<q; i++) {
	cin>>t[i]>>a[i];
    }

    auto bound = [&](ll val, ll l, ll r) {
	val = min(val, r);
	val = max(l, val);
	return val;
    };


    int ri=0;
    ll sgn = -1;
    ll lo = 0;
    ll hi = x;
    ll dx = 0;
    ll last = 0;
    
    for (int qi=0; qi<q; ) {
	if (ri==k || t[qi] < r[ri]) {
	    ll res = bound(dx+a[qi], lo, hi);
	    res += sgn*(t[qi]-last);
	    res = bound(res, 0, x);
	    cout<<res<<"\n";
	    qi++;
	    
	} else {
	    ll dr = sgn*(r[ri]-last);

	    lo = bound(lo+dr, 0, x);
	    hi = bound(hi+dr, 0, x);
	    dx += dr;

	    last = r[ri];
	    ri++;
	    sgn = -sgn;	    
	}
    }
    
    
    return 0;
}
