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 CPP):

Solution:
#include <bits/stdc++.h>
const char road[1000] = "C:\\data\\";
char roadnum[1000] = "";
char roadin[1000] = "";
char roadout[1000] = "";
void writeIntoFile(char num[]) {
  strcpy(roadnum, num);
  strcpy(roadin, road);
  strcat(roadin, roadnum);
  strcat(roadin, ".");
  strcpy(roadout, roadin);
  strcat(roadin, "in");
  strcat(roadout, "out");
  freopen(roadin, "r", stdin);
  freopen(roadout, "w", stdout);
}
void solve();
void solveM() {
  char ch[50];
  for (int i = 1; i <= 30; i++) {
    sprintf(ch, "%d", i);
    writeIntoFile(ch);
    solve();
  }
}
const long long N = 1e18;
const long long M = 1e5;
const long long MOD = 1e9;
long long mod, u, v;
long long p2[70];
void buildP2() {
  int i;
  p2[0] = 1;
  for (i = 1; i < 63; i++) {
    p2[i] = p2[i - 1] * 2;
  }
}
long long calSeg(long long f, long long tl, long long tr, int dep) {
  if (mod <= 1) return 0;
  long long la = f + (tr - tl) * p2[dep];
  if (v < f || u > la) return 0;
  long long inf, ine;
  long long a1, an;
  an = v;
  if (an > la) an = la;
  a1 = u - f;
  an = an - f;
  if (a1 < 0) a1 = 0;
  inf = a1 / p2[dep];
  if (a1 % p2[dep]) inf++;
  ine = an / p2[dep];
  long long ds = ine - inf + 1;
  long long sum = (ds) % mod * (f % mod) % mod;
  long long d;
  if (ds % 2 == 0) {
    ds /= 2;
    d = (ine % mod + (inf % mod)) % mod * (ds % mod) % mod * (p2[dep] % mod) %
        mod;
  } else {
    d = ((ine + inf) / 2) % mod * (ds % mod) % mod * (p2[dep] % mod) % mod;
  }
  sum = (sum + d) % mod;
  return sum;
}
long long queryTree(long long f, long long tl, long long tr, long long ql,
                    long long qr, int dep) {
  if (ql > tr || qr < tl) return 0;
  if (ql <= tl && tr <= qr) {
    return calSeg(f, tl, tr, dep);
  }
  long long mid = (tl + tr) / 2;
  long long sum = 0;
  sum += queryTree(f, tl, mid, ql, qr, dep + 1);
  sum %= mod;
  sum += queryTree(f + p2[dep], mid + 1, tr, ql, qr, dep + 1);
  sum %= mod;
  return sum;
}
void solve() {
  buildP2();
  long long n, m;
  long long l, r, sum;
  while (scanf("%lld%lld%lld", &n, &m, &mod) != EOF) {
    assert(1 <= n && n <= N);
    assert(1 <= m && m <= M);
    assert(1 <= mod && mod <= MOD);
    while (m--) {
      scanf("%lld%lld%lld%lld", &l, &r, &u, &v);
      assert(1 <= l && l <= r && r <= n);
      assert(1 <= u && u <= v && v <= N);
      sum = queryTree(1, 1, n, l, r, 0);
      printf("%lld\n", sum);
    }
  }
}
int main() {
  solve();
  return 0;
}