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>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;

int r[200000];
ll a[200000],b[200000],c[200000];
int main(){
	ll x,k;scanf("%lld%lld",&x,&k);
	rep(i,k)scanf("%d",&r[i+1]);
	c[0]=x;
	rep(i,k){
		int s=r[i+1]-r[i];
		if(i%2==0)s=-s;
		a[i+1]=min(max(a[i]+s,0LL),x);
		b[i+1]=b[i]+s;
		c[i+1]=min(max(c[i]+s,0LL),x);
	}
	int q;scanf("%d",&q);
	rep(i,q){
		int t,d;scanf("%d%d",&t,&d);
		int j=upper_bound(r,r+k+1,t)-r-1;
		int s=t-r[j];
		if(j%2==0)s=-s;
		ll na=min(max(a[j]+s,0LL),x);
		ll nb=b[j]+s;
		ll nc=min(max(c[j]+s,0LL),x);
		printf("%lld\n",min(max(nb+d,na),nc));
	}
}