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>
using namespace std;
typedef long long ll;
const int MN = 101000;
ll r[MN];
ll v[MN],w[MN],u[MN];
ll vt[MN],vb[MN],sn[MN];
int main() {
	ll x,k,q;
	scanf("%lld %lld ",&x,&k);
	r[0] = 0;
	for(int i=0;i<k;i++) {
		scanf("%lld ",&r[i+1]);
	}
	k++;
	sn[0] = -1;
	vt[0] = 0;vb[0] = 0;
	v[0] = x;w[0] = 0;u[0] = 0;
	for(int i=1;i<k;i++) {
		u[i] = u[i-1]+sn[i-1]*(r[i]-r[i-1]);
		v[i] = v[i-1]+sn[i-1]*(r[i]-r[i-1]);
		w[i] = w[i-1]+sn[i-1]*(r[i]-r[i-1]);
		vt[i] = vt[i-1];vb[i] = vb[i-1];
		if(v[i] > x) {
			vt[i] += v[i]-x;
			v[i] = x;
		}
		if(w[i] > x) {w[i] = x;}
		if(w[i] < 0) {
			vb[i] -= w[i];
			w[i] = 0;
		}
		if(v[i] < 0) {v[i] = 0;}
		sn[i] = sn[i-1]*-1;
	}
	scanf("%lld ",&q);
	while(q--) {
		ll t,a;
		scanf("%lld %lld ",&t,&a);
		int id = upper_bound(r,r+k,t)-r-1;
		//printf("%lld -> %lld, id %ld\n",w[id],v[id],id);
		if(a <= vb[id]) {
			a = w[id];
		} else if(a >= x-vt[id]) {
			a = v[id];
		} else {
			a += u[id];
		}
		a += sn[id]*(t-r[id]);
		a = max(a,0LL);
		a = min(a,x);
		printf("%lld\n",a);
	}
}
