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>
#define FOR(i,bg,ed) for(ll i=(bg);i<(ed);i++)
#define REP(i,n) FOR(i,0,n)
#define REP1(i,n) for(ll i=1;i<=(n);i++)
#define MOD 1000000007
#define int long long
#define all(c) (c).begin(), (c).end()
#define uniq(c) c.erase(unique(all(c)), (c).end())
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
const int INF = 1e9;

ll T;   //合計T[g]の砂
int K;  //砂時計をひっくり返す回数
int Q;  //クエリ数

signed main()
{
    cin >> T;
    cin >> K;
    set<P> events;
    REP(i,K) {
        int t;  //ひっくり返す時刻。元から昇順で並んでいる
        cin >> t;
        events.insert(P(t, -1));    //reverse query
    }
    cin >> Q;
    REP(i,Q) {
        ll t;   //時刻
        ll f;   //始めのAの砂の量
        cin >> t >> f;
        events.insert(P(t, f));
    }

    bool inc = false;   //A,Bどちらが上か。Aが上のときfalse
    ll pt = 0;
    ll a = 0;
    ll b = T;
    ll c = 0;
    for (P e : events) {
        ll t = e.first;     //ひっくり返す、もしくはクエリの時刻
        ll x = e.second;    //-1ならreverse query

        if (inc) {  //Bが上
            ll d = t - pt;
            c += d;
            if (b + c > T) b = T - c;
            if (a > b) a = b;
        } else {    //Aが上
            ll d = t - pt;
            c -= d;
            if (a + c < 0) a = -c;
            if (a > b) b = a;
        }
        if (x == -1) {  //reverse
            inc = !inc;
        } else {
            cout << max(a, min(b, x)) + c << endl;
        }
        pt = t;
    }
}
