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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(a);i>(b);i--)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll; typedef pair<ll, ll> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename A,size_t N,typename T>void Fill(A(&array)[N],const T &val){fill((T*)array, (T*)(array+N), val);}
const int inf = INT_MAX / 2; const ll INF = LLONG_MAX / 2;
//template end



int main(){
    int x,k; scanf("%d%d",&x,&k);
    vector<P> que;
    rep(i,0,k){
    	int r; scanf("%d",&r);
    	que.push_back({r,-1});
    }
    int q; scanf("%d",&q);
    rep(i,0,q){
    	int t,a; scanf("%d%d",&t,&a);
    	que.push_back({t,a});
    }
    sort(ALL(que)); vector<int> ans;
    int pre=0,minn=0,maxx=x,drc=-1,diff=0;
    rep(i,0,que.size()){
    	int dy=que[i].first-pre; diff+=drc*dy;
    	minn=min(max(minn,-diff),x-diff);
    	maxx=min(max(maxx,-diff),x-diff);
    	if(que[i].second==-1)drc*=-1;
    	else{
    		int val=que[i].second;
    		ans.push_back(min(max(val,minn),maxx)+diff);
    	} pre=que[i].first;
    }
    rep(i,0,q)printf("%d\n",ans[i]);
    return 0;
}
