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<(int)(n);i++)
#define ALL(x) (x).begin(),(x).end()
#define LL long long

using namespace std;

int main(){

    LL X;cin>>X;
    int K;cin>>K;
    LL r[K+1];REP(i,K)cin>>r[i+1];
    r[0]=0;

    LL sum[K+1];
    sum[0]=0;
    REP(i,K)sum[i+1]=sum[i]+(r[i+1]-r[i])*(i%2==0 ? 1:-1);
    pair<LL,LL> range[K+1];
    range[0]={0LL,X};

    REP(i,K){
        range[i+1].first=range[i].first;range[i+1].second=range[i].second;
        if(i%2==0)range[i+1].first=max(range[i+1].first,sum[i+1]);
        else range[i+1].second =min(range[i+1].second,X+sum[i+1]);
    }

    LL a0[K+1];a0[0]=0;
    LL ax[K+1];ax[0]=X;

    REP(i,K){
        a0[i+1]=a0[i]+(r[i+1]-r[i])*(i%2==0 ? -1:1);
        a0[i+1]=max(0LL,a0[i+1]);
        a0[i+1]=min(X,a0[i+1]);

    }

    REP(i,K){
        ax[i+1]=ax[i]+(r[i+1]-r[i])*(i%2==0 ? -1:+1);
        ax[i+1]=max(0LL,ax[i+1]);
        ax[i+1]=min(X,ax[i+1]);
    }

    int q;cin>>q;
    REP(i,q){
        LL t,a;cin>>t>>a;
        LL ans=0;
        int k = upper_bound(r,r+K+1,t)-r -1;
        if(a>range[k].first&&a<range[k].second)ans = a - sum[k];
        else if(a<=range[k].first)ans = a0[k];
        else ans = ax[k];

        ans += (t-r[k])*(k%2 ==0 ? -1:1);
        ans = max(0LL,ans);
        ans = min(X,ans);

        cout<<ans<<endl;

    }

    return 0;
}