AquaMoon and Chess

time limit per test: 1 second

memory limit per test: 256 megabytes

Articles

Cirno gave AquaMoon a chessboard of size $1×n$. Its cells are numbered with integers from $1$ to $n$ from left to right. In the beginning, some of the cells are occupied with at most one pawn, and other cells are unoccupied.

In each operation, AquaMoon can choose a cell ii with a pawn, and do either of the following (if possible):

• Move pawn from it to the $(i+2)$-th cell, if $i+2≤n$ and the $(i+1)$-th cell is occupied and the $(i+2)$-th cell is unoccupied.
• Move pawn from it to the $(i−2)$-th cell, if $i−2≥1$ and the $(i−1)$-th cell is occupied and the $(i−2)$-th cell is unoccupied.

You are given an initial state of the chessboard. AquaMoon wants to count the number of states reachable from the initial state with some sequence of operations. But she is not good at programming. Can you help her? As the answer can be large find it modulo $998244353$.

Input

The input consists of multiple test cases. The first line contains a single integer $t$ $(1≤t≤10000)$ — the number of test cases.

The first line contains a single integer $n (1≤n≤10^5)$ — the size of the chessboard.

The second line contains a string of $n$ characters, consists of characters “0” and “1”. If the $i$-th character is “1”, the ii-th cell is initially occupied; otherwise, the $i$-th cell is initially unoccupied.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output

For each test case, print the number of states that reachable from the initial state with some sequence of operations modulo $998244353$.

Example

Input

6
4
0110
6
011011
5
01010
20
10001111110110111000
20
00110110100110111101
20
11101111011000100010

Output

3
6
1
1287
1287
715

Note

In the first test case the strings “1100”, “0110” and “0011” are reachable from the initial state with some sequence of operations.

思路

字符0每个都是一个独立的个体，每两个连续的字符1是一个独立的个体，使用组合数即可计算总的状态数。（单独的字符1不影响组合数的计算）

代码

#include<bits/stdc++.h>

using namespace std;

const int maxn = 1e6+1;
const int mod = 998244353; 

long long M = 1, inv[maxn], mul[maxn], invmul[maxn];
long long C(long long n, long long m) {
    return mul[n] * invmul[m] % mod * invmul[n-m] % mod;
}
void init(int n) {
    inv[1] = invmul[0] = invmul[1] = mul[0] = mul[1] = 1;
    for(int i = 2; i <= n; ++ i) {
        inv[i] = (mod-mod/i) * inv[mod%i] % mod;
        invmul[i] = invmul[i-1] * inv[i] % mod;
        mul[i] = mul[i-1] * i % mod;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1; 
    cin >> T;
    init(maxn-1);
    while(T --) {
        int n, m = 0, N = 0, k = 0;
        cin >> n;
        string s;
        cin >> s;
        for(int i = 0; i < n; ++ i) {
            if (s[i] == '0') {
                N ++;
                m += k/2;
                k = 0;
            }
            else {
                k ++;
            }
        }
        m += k/2;
        cout << C(N+m, m) << "\n";
    }
    return 0;
}