Codeforces Round #475 (Div. 2) C. Alternating Sum（等比数列+逆元）

C. Alternating Sum

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two integers $a$ and $b$. Moreover, you are given a sequence $s_0,s_1,\dots ,s_n$. All values in $s$ are integers $1$ or $-1$. It's known that sequence is $k$-periodic and $k$ divides $n+1$. In other words, for each $k\le i\le n$ it's satisfied that $s_i=s_{i-k}$.

Find out the non-negative remainder of division of $\sum _{i=0}^n{s}_i{a}^{n-i}{b}^i$ by ${10}^9+9$.

Note that the modulo is unusual!

Input

The first line contains four integers $n,a,b$ and $k$ $(1\le n\le {10}^9,1\le a,b\le {10}^9,1\le k\le {10}^5)$.

The second line contains a sequence of length $k$ consisting of characters '+' and '-'.

If the $i$-th character (0-indexed) is '+', then $s_i=1$, otherwise $s_i=-1$.

Note that only the first $k$ members of the sequence are given, the rest can be obtained using the periodicity property.

Output

Output a single integer — value of given expression modulo ${10}^9+9$.

Examples

input

Copy

2 2 3 3
+-+

output

Copy

7

input

Copy

4 1 5 1
-

output

Copy

999999228

Note

In the first example:

$\left(\sum _{i=0}^n{s}_i{a}^{n-i}{b}^i\right)$ = $2^2{3}^0-2^1{3}^1+2^0{3}^2$ = 7

In the second example:

$\left(\sum _{i=0}^n{s}_i{a}^{n-i}{b}^i\right)=-1^4{5}^0-1^3{5}^1-1^2{5}^2-1^1{5}^3-1^0{5}^4=-781\equiv 999999228(\mathrm{mod}{10}^9+9)$.

$题意:\; 求解(\sum _{i=0}^n{s}_i{a}^{n-i}{b}^i)$ ，$s_i{\mathrm{要么是-要么是+，给出1-k的si,\; 对于 k\le i\le nk\le i\le n 则有si=si-ksi=si-k。}}^{}$

对于i, i+k, i + 2 *k.....可发现公比是(b/a)^k的等比数列，因为求和公式为a1(1-q^n)/（1-q）,(1-q^x)/（1-q）是一样的，所以计算出

1-k的(k∑i=0sian−ibi)在乘以(1-q^x)/（1-q）即可，项数x = (n+1)/k, 取模的时候，除法要逆元，用费马小定理即可。

#include <bits/stdc++.h>
#include <ext/hash_map>
#include <ext/hash_set>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
#define XINF INT_MAX
#define INF 0x3F3F3F3F
#define MP(X,Y) make_pair(X,Y)
#define PB(X) push_back(X)
#define REP(X,N) for(int X=0;X<N;X++)
#define REP2(X,L,R) for(int X=L;X<=R;X++)
#define DEP(X,R,L) for(int X=R;X>=L;X--)
#define CLR(A,X) memset(A,X,sizeof(A))
#define IT iterator
#define RIT reverse_iterator
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<int> VI;
typedef tree<PII, null_type, greater<PII>, rb_tree_tag, tree_order_statistics_node_update > rb_tree_set;
typedef tree<PII, PII, greater<PII>, rb_tree_tag, tree_order_statistics_node_update > rb_tree;
#define PQ std::priority_queue
#define HEAP __gnu_pbds::priority_queue
#define X first
#define Y second
#define lson(X) ((X)<<1)
#define rson(X) ((X)<<1|1)
#define N 110
#define M 200010
const ll  mod = 1e9 + 9;
ll q_pow(ll a, ll b)
{
	ll ans = 1;
	while(b) {
		if(b & 1)
			ans = ans * a % mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
int main()
{
	ll sum = 0, ans  = 0;
	ll a, b, k, n;
	scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
	getchar();
	char ch;
	ll head = q_pow(a, n);
	ll ni_a = q_pow(a, mod-2);
	REP(i, k) {
		scanf("%c", &ch);
		if(ch == '+')
			ans = (ans + head) % mod;
		else
			ans = (ans - head + mod) % mod;
		head = head * ni_a % mod * b % mod;
	}
	ll x = (n+1) / k;
	//ll q = q_pow(q_pow(a, k), mod-2) * q_pow(b, k) % mod;
	ll q = q_pow(b*ni_a%mod, k);
    if(q == 1) {
        printf("%lld\n", ans * x % mod);
    }else {
        sum = (1 - q_pow(q, x) + mod) % mod;    //负数取模.
        sum = sum * q_pow((1-q+mod)%mod, mod-2) % mod;
        printf("%lld\n", ans * sum % mod);
    }
	return 0;
}