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;

#define fst(t) std::get<0>(t)
#define snd(t) std::get<1>(t)
#define thd(t) std::get<2>(t)
#define unless(p) if(!(p))
#define until(p) while(!(p))

using ll = long long;
using P = std::tuple<int,int>;

const int dx[8] = {-1, 1, 0, 0, -1, -1, 1, 1}, dy[8] = {0, 0, -1, 1, -1, 1, -1, 1};

vector<tuple<int, int, int> > v;

int main(){
    std::cin.tie(nullptr);
    std::ios::sync_with_stdio(false);

    ll X, K;
    std::cin >> X >> K;

    for(int i=0;i<K;++i){
        int r;
        std::cin >> r;

        v.emplace_back(r, 2, -1);
    }

    int Q;
    std::cin >> Q;

    for(int i=0;i<Q;++i){
        int t, a;
        std::cin >> t >> a;

        v.emplace_back(t, 1, a);
    }

    sort(v.begin(), v.end());

    // 今 max{lb, min{ub, x + s}}
    ll lb = 0, ub = X, s = 0;
    int lastTime = 0, sgn = -1;
    for(const auto& p : v){
        int t, qt, a;
        tie(t, qt, a) = p;

        int passedTime = t - lastTime;
        if(qt == 1){
            ll x = max(lb, min<ll>(ub, a + s));
            x = max(0ll, min<ll>(X, x + sgn * passedTime));
            std::cout << x << std::endl;
        }else{
            lb = max(0ll, min<ll>(X, lb + sgn * passedTime));
            ub = max(0ll, min<ll>(X, ub + sgn * passedTime));
            s += sgn * passedTime;

            sgn *= -1;
            lastTime = t;
        }
    }
}
