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<stdio.h>
#include<map>
#include<cmath>
#include<stdlib.h>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>

using namespace std;

typedef pair<int,int> pii;
typedef long long ll;

const int MX = 100005;
const int MM = 998244353;

struct EV{
	int o, t, a, ad;
	EV(int o, int t, int a, int ad):o(o), t(t), a(a), ad(ad){}
	bool operator<(EV l)const{
		return t < l.t;
	}
};

int N;
int ans[MX], D[MX];

int main()
{
	vector<EV> L;
	int X, K, Q, t, a;
	scanf("%d%d", &X, &K);
	for(int i = 1; i <= K; i++){
		scanf("%d", D+i);
		L.emplace_back(0, D[i], 0, 0);
	}
	scanf("%d", &Q);
	for(int i = 1; i <= Q; i++){
		scanf("%d%d", &t, &a);
		L.emplace_back(1, t, a, i);
	}

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

	int l = 0, r = X, u = 0, ch = 0, b = 0;
	for(EV c : L){
		if( c.o == 0 ){
			int d = c.t - b; b = c.t;
			if( ch ){
				int v = X-(u + (r-l));
				r = max(l, r - max(0, d-v));
				u = min(X, u+d);
			}
			else{
				l = min(r, l + max(0, d - u));
				u = max(0, u-d);
			}
			ch = !ch;
		}
		else{
			int d = c.t - b, a = c.a;
			int v = u + min(r-l, max(0, a-l));
			if( !ch ){
				ans[c.ad] = max(0, v - d);
			}
			else ans[c.ad] = min(X, v + d);
		}
	}

	for(int i = 1; i <= Q; i++) printf("%d\n", ans[i]);
}
