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"

#pragma warning(disable:4996)
using namespace std;
using ld = long double;
const ld eps = 1e-9;




int main() {
	int X;cin>>X;
	int K;cin>>K;
	vector<int>rs(K+1);
	vector<int>as(K+1),bs(K+1,X);

	int pre=0;
	int a=0;
	int b=0;
	
	for (int i = 0; i < K; ++i) {
		cin>>rs[i+1];
		
	}
	vector<int>mins(K+1);
	vector<int>maxs(K+1,X);
	for (int i = 0; i < K; ++i) {
		int sa = rs[i+1]-rs[i];

		if (i % 2 == 0) {
			mins[i + 1] = max(0, mins[i] - sa);
			maxs[i+1]=max(0,maxs[i]-sa) ;
		}
		else {
			maxs[i + 1] = min(X, maxs[i] + sa);
			mins[i+1]=min(X,mins[i]+sa);
		}
	}
	for (int i = 0; i < K; ++i) {
		int sa = rs[i + 1] - rs[i];

		if (i % 2 == 0) {
			int yoyuu=mins[i];
			as[i+1]=as[i]+max(0,sa-yoyuu);
			bs[i+1]=bs[i];
			as[i+1]=min(as[i+1],bs[i+1]);
		}
		else {
			int yoyuu=X-maxs[i];
			as[i+1]=as[i];
			bs[i + 1] = bs[i]-max(0,sa-yoyuu);
			bs[i+1]=max(bs[i+1],as[i+1]);
		}
	}

	int Q;cin>>Q;
	while (Q--) {
		int t,a;cin>>t>>a;
		auto it=prev(upper_bound(rs.begin(),rs.end(),t));

		int x=it-rs.begin();
		int dif=t-*it;

		bool flag=x%2;

		int ans=-1;

		int d;
		if (a < as[x]) {
			d=mins[x];
		}
		else if (a < bs[x]) {
			d=mins[x]+(a-as[x]);
		}
		else {
			d=maxs[x];
		}
		ans=x%2?(d+dif):(d-dif);
		ans=max(0,min(X,ans));
		cout<<ans<<endl;
	}

	return 0;
} 