Dreamoon likes sequences very much. So he created a problem about the sequence that you can't find in OEIS:
You are given two integers d,md,m, find the number of arrays aa, satisfying the following constraints:
- The length of aa is nn, n≥1n≥1
- 1≤a1<a2<⋯<an≤d1≤a1<a2<⋯<an≤d
- Define an array bb of length nn as follows: b1=a1b1=a1, ∀i>1,bi=bi−1⊕ai∀i>1,bi=bi−1⊕ai, where ⊕⊕ is the bitwise exclusive-or (xor). After constructing an array bb, the constraint b1<b2<⋯<bn−1<bnb1<b2<⋯<bn−1<bn should hold.
Since the number of possible arrays may be too large, you need to find the answer modulo mm.
Input
The first line contains an integer tt (1≤t≤1001≤t≤100) denoting the number of test cases in the input.
Each of the next tt lines contains two integers d,md,m (1≤d,m≤1091≤d,m≤109).
Note that mm is not necessary the prime!
Output
For each test case, print the number of arrays aa, satisfying all given constrains, modulo mm.
Example
input
Copy
10 1 1000000000 2 999999999 3 99999998 4 9999997 5 999996 6 99995 7 9994 8 993 9 92 10 1
output
Copy
1 3 5 11 17 23 29 59 89 0
#include <iostream> #include <vector> #include <algorithm> #include <string> #include <set> #include <queue> #include <map> #include <sstream> #include <cstdio> #include <cstring> #include <numeric> #include <cmath> #include <iomanip> #include <deque> #include <bitset> //#include <unordered_set> //#include <unordered_map> //#include <bits/stdc++.h> //#include <xfunctional> #define ll long long #define PII pair<int, int> #define rep(i,a,b) for(int i=a;i<=b;i++) #define dec(i,a,b) for(int i=a;i>=b;i--) #define pb push_back #define mk make_pair using namespace std; int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } }; const long long INF = 0x7f7f7f7f7f7f7f7f; const int inf = 0x3f3f3f3f; const double pi = 3.14159265358979; const int mod = 998244353; const int N = 2e5+5; //if(x<0 || x>=r || y<0 || y>=c) inline ll read() { ll x = 0; bool f = true; char c = getchar(); while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); } while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar(); return f ? x : -x; } ll gcd(ll m, ll n) { return n == 0 ? m : gcd(n, m%n); } int main() { vector<ll> a(33,1); for (int i = 1; i <= 32; i++) { a[i] = a[i - 1] * 2; } int T; cin >> T; while (T--) { ll d, m; cin >> d >> m; ll n=upper_bound(a.begin(), a.end(), d) - a.begin()-1; ll res=1; for (int i = 0; i < n; i++) { res *= (a[i] + 1)%m; res%=m; } res = res*(d - a[n] + 2) - 1; res = (res+m) % m; cout << res << endl; } return 0; }