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>
//#include <math.h>
using namespace std;

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define BIT(x,n) bitset<n>(x)
#define PI 3.14159265358979323846

typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> PP;

struct edge {
	int to, cost;
	edge(int t,int c):to(t),cost(c) {}
};
int dx[]={1,-1,0,0},dy[]={0,0,1,-1};

//-----------------------------------------------------------------------------

int X,A[100000],R[100000],T[100000];
int K,Q;
int dir=-1,l,u,m;
int t,dif,j;

void calc(int tmp) {
	t+=abs(tmp),dif+=tmp;
	l=max(0,min(X,l+tmp));
	u=max(0,min(X,u+tmp));
}

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

	cin>>X>>K;
	REP(i,K) cin>>R[i];
	cin>>Q;
	REP(i,Q) cin>>T[i]>>A[i];

	u=X;
	REP(i,Q) {
		while(j<K&&R[j]<T[i]) calc((R[j++]-t)*dir),dir=-dir;
		calc((T[i]-t)*dir);
		cout<<max(l,min(u,A[i]+dif))<<endl;
	}

	return 0;
}
