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 MAX=2e5+5;
const int INF=0x3f3f3f3f;
template<class T>bool gmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>bool gmin(T &a,T b){return a>b?a=b,1:0;}

ll n,q,t,a,r[MAX],X;
int main()
{
    scanf("%lld%lld",&X,&n);
    for(int i=1;i<=n;i++)scanf("%lld",&r[i]);
    scanf("%lld",&q);
    ll minsand=0,maxsand=X,val=0,k=1,sign=-1;
    while(q--)
    {
        scanf("%lld%lld",&t,&a);
        while(k<=n&&r[k]<=t)
        {
            ll v=sign*(r[k]-r[k-1]);
            minsand+=v;
            maxsand+=v;
            val+=v;
            gmax(minsand,0ll);
            gmin(minsand,(ll)X);
            gmax(maxsand,0ll);
            gmin(maxsand,(ll)X);
            sign*=-1;
            k++;
        }
        ll now=val+a;
        gmax(now,minsand);
        gmin(now,maxsand);
        now+=sign*(t-r[k-1]);
        gmax(now,0ll);
        gmin(now,(ll)X);
        printf("%lld\n",now);
    }
}
