K-Calculator

Do you know kimchi?

Do you know Lee Guk-ryeol?

Do you know calculators? Perfect

Guk-ryeol has set a calculator problem three times already, and now he has set another one. Let us see how long this keeps going. You have run into yet another calculator problem, which is annoying, but to win the prize you have to evaluate the given expression according to the rules.

Guk-ryeol gave expressions in the same format three times, and since Guk-ryeol, the contestants, and the reviewers all do not want to suffer through parsing, he changed the format. In this input, numbers and operators alternate, with a space between each number and operator. If we write the $i$-th number of the expression as $X_i$ and the $i$-th operator as $Op_i$, an expression with $N$ numbers can be written as $X_1$ $Op_1$ $X_2$ $Op_2$ ... $Op_{N-1}$ $X_N$. The operators are +, -, *, /. The input never ends with an operator, never has a negative $X_i$, and never has unnecessary leading zeros. That is, cases such as $-1-1$, $2+-3$, and $001+0002$ are not given as input.

An expression made of $N$ numbers and $N-1$ positive integers $u_i$ are given. You evaluate it according to the following rules.

1. Repeat the following $N-1$ times.
2. Take the most recently computed result, written as the reduced fraction $P/Q$. Let $Q^{-1}$ be the multiplicative inverse of $Q$ modulo $10^9+7$, and let $v$ be the remainder of $P \times Q^{-1}$ modulo $10^9+7$. If there is no previously computed result, $v=0$.
3. Evaluate the $(u_i \oplus v)$-th operator and output the result of that evaluation. Here $\oplus$ means bitwise XOR.
4. Remove the operator and operands you evaluated, and put the result in their place.

For example, given the expression $3 - 2 \times 5 + 10$ and the positive integers $u = [1,3,14]$, you evaluate as follows.

1. Since $1 \oplus 0 = 1$, evaluate the 1st operator. The result is $3-2=1$, so output $1$, and the expression becomes $1 \times 5 + 10$.
2. Since $3 \oplus 1 = 2$, evaluate the 2nd operator. The result is $5+10 =15$, so output $15$, and the expression becomes $1 \times 15$.
3. Since $14 \oplus 15 = 1$, evaluate the 1st operator. The result is $1 \times 15 =15$, so output $15$.

Division in this problem is real division. That is, $5/2$ is $2.5$, not $2$.

Write a program that evaluates an expression this way.

The first line gives $N$, the number of operands. ($2 \le N \le 500\,000$)

The second line gives an expression made only of operands and +, -, *, /. Operands and operators are separated by spaces, and each operand is an integer between $0$ and $10^9+6$ inclusive.

The third line gives $u_1$, ..., $u_{N-1}$, separated by spaces. ($1 \le u_i \oplus v \le N - i$)

No case in the evaluation divides by $0$.

For each $i$, write the result for $u_i$ as the reduced fraction $P/Q$, let $Q^{-1}$ be the multiplicative inverse of $Q$ modulo $10^9+7$, and print the remainder of $P \times Q^{-1}$ modulo $10^9+7$ on the $i$-th line.