## SPP - Recursive Sequence (Version II)

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Sequence (ai) of natural numbers is defined as follows:

ai = bi (for i <= k)
ai = c1ai-1 + c2ai-2 + ... + ckai-k (for i > k)

where bj and cj are given natural numbers for 1<=j<=k. Your task is to compute am + am+1 + am+2 + ... + an for given m <= n and output it modulo a given positive integer p.

### Input

On the first row there is the number C of test cases (equal to about 50).
Each test contains four lines:
k - number of elements of (c) and (b) (1 <= k <= 15)
b1, ... bk - k natural numbers where 0 <= bj <= 109 separated by spaces.
c1, ... ck - k natural numbers where 0 <= cj <= 109 separated by spaces.
mnp - natural numbers separated by spaces (1 <= m <= n <= 10181<= p <= 108).

### Output

Exactly C lines, one for each test case: (am + am+1 + am+2 + ... + an) modulo p.

### Example

```Input:
1
2
1 1
1 1
2 10 1000003

Output:
142```

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>
#include <limits.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;

// #define MOD (ll)1000000007
#define pb   push_back
#define EPS 1e-9
#define FOR(i,n)  for(int i = 0;i < n; i++)
#define FORE(i,a,b) for(int i = a;i <= b; i++)
#define pi(a)   printf("%d\n", a)
#define all(c)  c.begin(), c.end()
#define tr(container, it)   for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define gc getchar_unlocked
#define sdi(a, b)   si(a);si(b)
#define io ios_base::sync_with_stdio(false);cin.tie(NULL);
#define endl '\n'
#define F first
#define S second
#define FILL(arr, val)  memset(arr, val, sizeof(arr))

template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}

void si(int &x){
register int c = gc();
x = 0;
int neg = 0;
for(;((c<48 || c>57) && c != '-');c = gc());
if(c=='-') {neg=1;c=gc();}
for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
if(neg) x=-x;
}
ll MOD;
struct Matrix{
vector< vector<int> > mat; // the contents of matrix as a 2D-vector
int n_rows, n_cols; // number of rows and columns
Matrix() {}
Matrix(vector< vector<int> > values): mat(values), n_rows(values.size()),
n_cols(values[0].size()) {}

static Matrix identity_matrix(int n){
// Return I_n - the identity matrix of size n.
// This function is static, because it creates a new Matrix instance
vector< vector<int> > values(n, vector<int>(n, 0));
for(int i = 0; i < n; i++)
values[i][i] = 1;
return values;
}

Matrix operator*(const Matrix &other) const {
ll MOD2 = MOD*MOD;
int n = n_rows, m = other.n_cols;
vector <  vector<int> > result(n_rows, vector<int>(n_cols, 0));
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++) {
long long tmp = 0;
for(int k = 0; k < n_cols; k++) {
tmp += mat[i][k] * 1ll * other.mat[k][j];
while(tmp >= MOD2)
tmp -= MOD2;
}
result[i][j] = tmp % MOD;
}
// Multiply matrices as usual, then return the result as the new Matrix
return Matrix(result);
}

inline bool is_square() const{
return n_rows == n_cols;
}
};

Matrix M_powers[55];

void precalc_powers(Matrix M)
{
assert(M.is_square());
M_powers[0] = M;

for(int i = 1; i < 55; i++)
M_powers[i] = M_powers[i - 1] * M_powers[i - 1];
}

Matrix fast_exponentiation_with_precalc(int power)
{
Matrix result = Matrix::identity_matrix(M_powers[0].mat.size());
int pointer = 0;
while(power) {
if(power & 1)
result = result * M_powers[pointer];
pointer++;
power >>= 1;
}
return result;
}

int main(){
// io;
int t;
si(t);
while(t--){
int k;
si(k);
int b[k];
FOR(i,k)	si(b[i]);
int c[k];
FOR(i,k)	si(c[i]);
ll m, n;
scanf("%lld%lld%lld",&m,&n,&MOD);
if(n <= k){
ll sum1 = 0;
for(int i = 0;i < n; i++){
sum1+=b[i];
}
ll sum2 = 0;
for(int i = 0;i < m-1; i++){
sum2+=b[i];
}
ll ans = sum1-sum2;
if(ans < 0)
ans = (ans+MOD)%MOD;
printf("%lld\n", ans%MOD);
}
//M^(n-k)
else{
vector<vector<int> > values(k+1, vector<int>(k+1, 0));
ll sum = 0;
for(int i = 0;i < k; i++){
sum += b[i];
}
values[0][0] = 1;
for(int i = 1;i <= k; i++){
values[i][0] = c[i-1];
values[i][1] = c[i-1];
}
for(int i = 1;i <k; i++){
values[i][i+1] = 1;
}
Matrix M(values);
precalc_powers(M);
vector<vector<int> > values2(1, vector<int>(k+1, 0));
values2[0][0] = sum;
for(int i = 1;i <= k; i++){
values2[0][i] = b[k-i];
}
Matrix M2(values2);
ll ans1 = (M2*fast_exponentiation_with_precalc(n-k)).mat[0][0];
ll sum2 = 0, ans2;
if(m <= k){
for(int i = 0;i < m-1; i++){
sum2+=b[i];
}
ans2 = sum2;
}else
ans2 = (M2*fast_exponentiation_with_precalc(m-k-1)).mat[0][0];
ll ans = ans1-ans2;
if(ans < 0)
ans = (ans+MOD)%MOD;
printf("%lld\n", ans);
}
}
return 0;
}``````
Copy The Code &

Input

cmd
1
2
1 1
1 1
2 10 1000003

Output

cmd
142
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## Demonstration

SPOJ Solution-Recursive Sequence (Version II)-Solution in C, C++, Java, Python