上来一看就是维护一个代数结构
以为要来个线段树 瞬间不想写了
结果发现每次都是整个区间操作 即只对线段树根节点操作
想存double水一下结果数据有点强 水不过去
所以就写个逆元就好啦
感觉这个还没上个难

typedef long long LL;

const LL mod = 1e9 + 7;

class Fancy {
public:
    LL mul = 1, add = 0;
    vector<LL> nums;
    Fancy() {
        mul = 1, add = 0;
        nums.clear();
    }
    LL qmi(LL a, LL b){
        LL res = 1;
        a %= mod;
        while(b){
            if(b & 1) (res *= a) %= mod;
            a = a % mod * a % mod;
            b >>= 1;
        }
        return res;
    }
    void append(int val) {
        nums.push_back((val - add) % mod * qmi(mul, mod - 2) % mod);
    }

    void addAll(int inc) {
        (add += inc) %= mod;
    }

    void multAll(int m) {
        (mul *= m) %= mod;
        (add *= m) %= mod;
    }

    int getIndex(int idx) {
        if(idx >= nums.size()) return -1;
        return (nums[idx] % mod * mul % mod + add + mod) % mod;
    }
};

/**
 * Your Fancy object will be instantiated and called as such:
 * Fancy* obj = new Fancy();
 * obj->append(val);
 * obj->addAll(inc);
 * obj->multAll(m);
 * int param_4 = obj->getIndex(idx);
 */