Problem: Let a be an array consisting of n numbers. The array's elements are numbered from 1 to n, even is an array consisting of the numerals whose numbers are even in a (eveni = a2i, 1 ≤ 2i ≤ n), odd is an array consisting of the numberals whose numbers are odd in а (oddi = a2i - 1, 1 ≤ 2i - 1 ≤ n). Then let's define the transformation of array F(a) in the following manner:

  * if n > 1, F(a) = F(odd) + F(even), where operation " + " stands for the arrays' concatenation (joining together) 
  * if n = 1, F(a) = a



Let a be an array consisting of n numbers 1, 2, 3, ..., n. Then b is the result of applying the transformation to the array a (so b = F(a)). You are given m queries (l, r, u, v). Your task is to find for each query the sum of numbers bi, such that l ≤ i ≤ r and u ≤ bi ≤ v. You should print the query results modulo mod.

Input

The first line contains three integers n, m, mod (1 ≤ n ≤ 1018, 1 ≤ m ≤ 105, 1 ≤ mod ≤ 109). Next m lines describe the queries. Each query is defined by four integers l, r, u, v (1 ≤ l ≤ r ≤ n, 1 ≤ u ≤ v ≤ 1018).

Please do not use the %lld specificator to read or write 64-bit integers in C++. Use %I64d specificator.

Output

Print m lines each containing an integer — remainder modulo mod of the query result.

Examples

Input

4 5 10000
2 3 4 5
2 4 1 3
1 2 2 4
2 3 3 5
1 3 3 4


Output

0
5
3
3
3


Input

2 5 10000
1 2 2 2
1 1 4 5
1 1 2 5
1 1 1 3
1 2 5 5


Output

2
0
0
1
0

Note

Let's consider the first example. First let's construct an array b = F(a) = F([1, 2, 3, 4]). 

  * Step 1. F([1, 2, 3, 4]) = F([1, 3]) + F([2, 4])
  * Step 2. F([1, 3]) = F([1]) + F([3]) = [1] + [3] = [1, 3]
  * Step 3. F([2, 4]) = F([2]) + F([4]) = [2] + [4] = [2, 4]
  * Step 4. b = F([1, 2, 3, 4]) = F([1, 3]) + F([2, 4]) = [1, 3] + [2, 4] = [1, 3, 2, 4]

Thus b = [1, 3, 2, 4]. Let's consider the first query l = 2, r = 3, u = 4, v = 5. The second and third positions in the array b do not have numbers in the range [4, 5], so the sum obviously equals zero. Let's consider the second query l = 2, r = 4, u = 1, v = 3. The second and third positions have two numbers that belong to the range [1, 3], their sum equals 5.

Solution (in JAVA):

Solution:

import java.util.*;
import java.io.PrintWriter;
import java.math.*;
import static java.lang.Character.isDigit;
import static java.lang.Character.isLowerCase;
import static java.lang.Character.isUpperCase;
import static java.lang.Math.*;
import static java.math.BigInteger.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;

public class Main{
    static void debug(Object...os){
//      System.err.println(deepToString(os));
    }
    
    void run(){
        long n=nextLong();
        int m=nextInt();
        int mod=nextInt();
        long[] ls=new long[m];
        long[] rs=new long[m];
        long[] us=new long[m];
        long[] vs=new long[m];
        for(int i=0;i<m;i++) {
            ls[i]=nextLong();rs[i]=nextLong();us[i]=nextLong();vs[i]=nextLong();
        }
        StringBuilder sb = new StringBuilder();
        PrintWriter out = new PrintWriter(System.out);
        for(int i=0;i<m;i++) {
            debug("i",i);
            long res = 0;
            res += calc(n,mod,rs[i],vs[i]);
            debug(res);
            res -= calc(n,mod,ls[i]-1,vs[i]);
            debug(res);
            res -= calc(n,mod,rs[i],us[i]-1);
            debug(res);
            res += calc(n,mod,ls[i]-1,us[i]-1);
            debug(res);
            res %= mod;
            res += mod;
            res %= mod;
            sb.append(res);
            sb.append('\n');
            if(sb.length()>100000) {
                out.print(sb);
                sb.setLength(0);
            }
        }
        out.print(sb);
        out.flush();
    }
    
    private long calc(long n,int mod,long r,long v){
        if(r<=0)return 0;
        if(v<=0)return 0;
        long[]res= dfs(n,mod,r,v);
        return res[0];
    }
    long[] dfs(long n,int mod,long r,long v) {
        if(r>n)throw new RuntimeException();
        if(r==0)return make(0,0);
        if(v<=0)return make(0,0);
        v=min(v,n);
        if(r==n) {
            v%=mod;
            return make(v*(v+1)/2%mod,v);
        }
        long left=(n+1)/2;
        long right=n/2;
        long[] ls = dfs(left,mod,min(left,r),(v+1)/2);
        long lval = (ls[0]*2-ls[1])%mod;
        if(lval>=mod)lval-=mod;
        long[] rs = dfs(right,mod,max(0,r-left),v/2);
        long rval = (rs[0]*2)%mod;
        long res = lval+rval;
        if(res>=mod)res-=mod;
        return make(res,ls[1]+rs[1]);
    }

    private long[] make(long a,long b){
        return new long[] {a,b};
    }

    int nextInt(){
        try{
            int c=System.in.read();
            if(c==-1) return c;
            while(c!='-'&&(c<'0'||'9'<c)){
                c=System.in.read();
                if(c==-1) return c;
            }
            if(c=='-') return -nextInt();
            int res=0;
            do{
                res*=10;
                res+=c-'0';
                c=System.in.read();
            }while('0'<=c&&c<='9');
            return res;
        }catch(Exception e){
            return -1;
        }
    }
    
    long nextLong(){
        try{
            int c=System.in.read();
            if(c==-1) return -1;
            while(c!='-'&&(c<'0'||'9'<c)){
                c=System.in.read();
                if(c==-1) return -1;
            }
            if(c=='-') return -nextLong();
            long res=0;
            do{
                res*=10;
                res+=c-'0';
                c=System.in.read();
            }while('0'<=c&&c<='9');
            return res;
        }catch(Exception e){
            return -1;
        }
    }
    
    double nextDouble(){
        return Double.parseDouble(next());
    }
    
    String next(){
        try{
            StringBuilder res=new StringBuilder("");
            int c=System.in.read();
            while(Character.isWhitespace(c))
                c=System.in.read();
            do{
                res.append((char)c);
            }while(!Character.isWhitespace(c=System.in.read()));
            return res.toString();
        }catch(Exception e){
            return null;
        }
    }
    
    String nextLine(){
        try{
            StringBuilder res=new StringBuilder("");
            int c=System.in.read();
            while(c=='\r'||c=='\n')
                c=System.in.read();
            do{
                res.append((char)c);
                c=System.in.read();
            }while(c!='\r'&&c!='\n');
            return res.toString();
        }catch(Exception e){
            return null;
        }
    }
    public static void main(String[] args){
        new Main().run();
    }
}