HDU

给定一个字符串，问其能通过重新排列生成多少不同的回文串。

显然只需要考虑一边就好了，考虑一边实际上就是多重集的全排列问题。

ll f[505];

void init() {
    f[0] = 1ll;
    for (int i = 1; i <= 500; i++) f[i] = f[i - 1] * i % MOD;
}


char s[1005];
map<char, int> mp;
int main() {
    int T = readint();
    init();
    while (T--) {
        mp.clear();
        scanf("%s", s);
        int len = strlen(s);
        for (int i = 0; i < len; i++) mp[s[i]]++;
        int cnt = 0;
        vector<int> v;
        for (auto it = mp.begin(); it != mp.end(); it++) if (it->second & 1) cnt++,v.push_back((it ->second  - 1) >> 1); else v.push_back(it->second / 2);
        if (cnt > 1) {
            puts("0");
            continue;
        }
        ll sum = 0;
        for (int i = 0; i < v.size(); i++) sum += v[i];
        ll res = f[sum];
        for (int i = 0; i < v.size(); i++) res *= ksm(f[v[i]], MOD - 2, MOD), res %= MOD;
        Put(res);
        puts("");
    }
}